logarithm problem solving examples

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Logarithmic Word Problems

Log Probs Expo Growth Expo Decay

What are logarithm word problems?

Logarithmic word problems, in my experience, generally involve either evaluating a given logarithmic equation at a given point, or else solving an equation for a given variable; they're pretty straightforward.

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What real-world problems use logarithms?

The classic real-world contexts for logarithm word problems are the measurement of acidity or alkalinity (that is, the measurement of pH), the measurement of sound (in decibels, or dB), and the measurement of earthquake intensity (on the Richter scale), among other uses ( link ).

Note: While log-based word problems are, in my experience, pretty straightforward, their statements tend to be fairly lengthy. Expect to have to plow through an unusual amount of text before they get to the point.

  • Chemists define the acidity or alkalinity of a substance according to the formula pH =  −log[H + ] where [H + ] is the hydrogen ion concentration, measured in moles per liter. Solutions with a pH value of less than 7 are acidic; solutions with a pH value of greater than 7 are basic; solutions with a pH of 7 (such as pure water) are neutral.

a) Suppose that you test apple juice and find that the hydrogen ion concentration is [H + ] = 0.0003 . Find the pH value and determine whether the juice is basic or acidic.

b) You test some ammonia and determine the hydrogen ion concentration to be [H + ] = 1.3 × 10 −9 . Find the pH value and determine whether the ammonia is basic or acidic.

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In each case, I need to evaluate the pH function at the given value of [H + ] . In other words, this exercise, despite all the verbiage, is just plug-n-chug.

Since no base is specified, I will assume that the base for this logarithm is 10 , so that this is the so-called "common" log. (I happen to know that 10 is indeed the correct base, but they should have specified.)

a) In the case of the apple juice, the hydrogen ion concentration is [H + ] = 0.0003 , so:

pH = −log[H + ]

= −log[0.0003]

= 3.52287874528...

This value is less than 7 , so the apple juice is acidic.

b) In the case of the ammonia, the hydrogen ion concentration is [H + ] = 1.3 × 10 −9 , so:

= −log[1.3 × 10 −9 ] = 8.88605664769...

This value is more than 7 , so the ammonia is basic.

(a) The juice is acidic with a pH of about 3.5 , and (b) the ammonia is basic with a pH of about 8.9 .

When a logarithm is given without a base being specified, different people in different contexts will assume different bases; either 10 , 2 , or e . Ask now whether or not bases will be specified for all exercises, or if you're going to be expected to "just know" the bases for certain formulas, or if you're supposed to "just assume" that all logs without a specified base have a base of... [find out which one].

  • "Loudness" is measured in decibels (abbreviated as dB). The formula for the loudness of a sound is given by dB = 10×log[I ÷ I 0 ] where I 0 is the intensity of "threshold sound", or sound that can barely be perceived. Other sounds are defined in terms of how many times more intense they are than threshold sound. For instance, a cat's purr is about 316 times as intense as threshold sound, for a decibel rating of:

dB = 10×log[I ÷ I 0 ]     = 10×log[ (316 I 0 ) ÷ I 0 ]     = 10×log[ 316 ]     = 24.9968708262...

...about 25 decibels.

Considering that prolonged exposure to sounds above 85 decibels can cause hearing damage or loss, and considering that a gunshot from a .22 rimfire rifle has an intensity of about I = (2.5 × 10 13 )I 0 , should you follow the rules and wear ear protection when practicing at the rifle range?

I need to evaluate the decibel equation at I = (2.5 × 10 13 )I 0 :

dB = 10log [ I ÷ I 0 ]     = 10log[ (2.5 ×10 13 )I 0 ÷ I 0 ]     = 10log[2.5 ×10 13 ]     = 133.979400087...

In other words, the squirrel gun creates a noise level of about 134 decibels. Since this is well above the level at which I can suffer hearing damage,

I should follow the rules and wear ear protection.

Algebra Tutors

  • Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by R = log[ I ÷ I 0  ] where I 0 is the "threshold quake", or movement that can barely be detected, and the intensity I is given in terms of multiples of that threshold intensity.

You have a seismograph set up at home, and see that there was an event while you were out that had an intensity of I = 989I 0 . Given that a heavy truck rumbling by can cause a microquake with a Richter rating of 3 or 3.5 , and "moderate" quakes have a Richter rating of 4 or more, what was likely the event that occurred while you were out?

To determine the probable event, I need to convert the intensity of the mystery quake into a Richter rating by evaluating the Richter function at I = 989I 0 :

R = log[ I ÷ I 0 ]     = log[ 989I 0 ÷ I 0 ]     = log[989]     = 2.9951962916...

A Richter rating of about 3 is not high enough to have been a moderate quake.

The event was probably just a big truck.

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A logarithm is the inverse of the exponential function . Specifically, a logarithm is the power to which a number (the base) must be raised to produce a given number.

For example, \(\log_2 64 = 6,\) because \( 2^6 = 64.\) In general, we have the following definition:

\( z \) is the base-\(x\) logarithm of \(y\) if and only if \( x^z = y \). In typical notation \[ \log_x y = z \iff x^z = y.\]

Properties of Logarithms - Basic

Worked examples using properties, properties of logarithms - intermediate, problem solving - basic, problem solving - intermediate, problem solving - advanced, applications.

First, we must know the basic structure of a logarithm \((\)abbreviated \(\log\) for convenience\().\) \(\log_a{b}=c\) can be rewritten as \(a^c=b,\) where \(a\) is called the base , \(c\) the exponent , and \(b\) the argument . Also, \(\log\) without a base is shorthand for the common \(\log\) of base \(10.\) Now that we know this, we can manipulate logs:

Other properties can be derived from these basic ones, especially when noting that these properties are inversable.

Simplify \(\log_2 \left(\dfrac{32}{9}\right)^2\) as much as possible. Try to follow the steps and identify what properties were used: \[\begin{align} \log_2 \left(\dfrac{32}{9}\right)^2 &=2 \cdot \log_2 \left(\frac{32}{9}\right)\\ &=2 \cdot ( \log_2 32 - \log_2 9)\\ &=2 \cdot \left( \log_2 2^5 -\log_2 3^2\right)\\ &=2 \cdot ( 5 \cdot \log_2 2 - 2 \cdot \log_2 3)\\ &=2 \cdot ( 5 \cdot 1 -2 \cdot \log_2 3)\\ &=10- 4 \log_2 3. \end{align}\] Note: \(\log_2 3\) can't be simplified further. Line 1 used the second property, line 2 put thingies into exponential form, line 3 used the third property, and lines 4 and 5 did basic simplification. \(_\square\)
Simplify \[\displaystyle{2\log_4{\sqrt{5}}+\frac{1}{2}\log_2{625}-\log_2{\frac{1}{5}}}.\] Again, try to follow the steps of the solution: \[\begin{align} 2\log_{(2^2)}{\big(5^\frac{1}{2}\big)}+\frac{1}{2}\log_2{\big(5^4\big)}-\log_2{\big(5^{-1}\big)} &=2\frac{\log 5^{\frac{1}{2}}}{\log 2^2}+\frac{1}{2}(4)(\log_2{5})+\log_2{5}\\ &=2\frac{\frac{1}{2}\log 5}{2\log 2}+2\log_2{5}+\log_2{5}\\ &=\frac{1}{2}\frac{\log 5}{\log 2}+3\log_2{5}\\ &=\frac{1}{2}\log_2{5}+3\log_2{5}\\ &=\frac{7}{2}\log_2{5}. \end{align}\] The first line shows that it is (usually) best to convert numbers so that they are integers to a power. Note that lines 4 reverses the process of the fourth property. \(_\square\)

\[ \] \(1.~\log _{ a }{ a } =1\)

Find the value of \(\log _{ 4 }{ 4 }.\) Using the property \(\log _{ a }{ a }=1,\) we get \(\log_{ 4 }{ 4 } =1. \ _\square\)

\[ \] \(2.~\log _{ a }{ (b^c) } =c\log _{ a }{ b } \)

Find the value of \(\log _{ 2 }{ 16 }.\) We have \[\begin{align} \log _{ 2 }{ 16 } &= \log _{ 2 }{ { 2 }^{ 4 } } &&\qquad \big(16={ 2 }^{ 4 }\big)\\ &=4\log _{ 2 }{ 2 } &&\qquad \big(\log { { a }^{ b } } = b\log { a } \big)\\ &= 4. \ _\square &&\qquad (\text{by property 1}) \end{align}\]

\[ \] \(3.~\log _{ a }{ (b \times c) } = \log _{ a }{ b }+ \log _{ a }{ c } \)

Find the value of \(\log { 90 }\) assuming \(\log { 3 } =0.47\). We have \[\begin{align} \log { 90 } &= \log { (9\times 10) } &&\qquad (90= 9 \times 10)\\ &=\log { 9 } + \log { 10 } &&\qquad \big(\log _{ a }{ (b\times c) } =\log _{ a }{ b } +\log _{ a }{ c } \big)\\ &=2\log { 3 } +1 &&\qquad \text{(by properties 2 and 1)}\\ &=2\times 0.47+1\\ &=0.94+1\\ &=1.94. \ _\square \end{align}\]

\[ \] \(4.~\displaystyle{\log _{ a }{ \frac { b }{ c } } = \log _{ a }{ b } - \log _{ a }{ c }}\)

Evaluate \(\log { 0.27 }\) assuming \(\log { 3 } =0.47\). We have \[\begin{align} \log { 0.27 } &= \log { \frac { 27 }{ 100 } } \\ &= \log { 27 } - \log { 100 } \\ &=3\log{ 3 } - 2\\ &=1.41 - 2\\ &=-0.59. \ _\square \end{align}\]

\[ \] \(5.\) \(\displaystyle{\log_{ a }{ b } = \frac { \log_{ c }{ b } }{ \log_{ c }{ a } }} \)

Find the value of \(\log_{ 32 }{ 2 }\). We have \[\begin{align} \log_{ 32 }{ 2 } &=\frac { \log_{ 2 }{ 2 } }{ \log_{ 2 }{ 32 } } \\ &=\frac { 1 }{ 5\log_{ 2 }{ 2 } } \\ &=\frac { 1 }{ 5 }\\ &={ 0.2 }. \ _\square \end{align} \]
What is the value of \( \log_3 15 + \log_3 81 - \log_3 5 ?\) Using the properties of logarithms, we can rewrite the given expression as follows: \[ \begin{align} \log_3 15 + \log_3 81 - \log_3 5 &= \log_3 15 - \log_3 5 + \log_3 3^4 \\ &= \log_3 \frac{15}{5} + \log_3 3^4 \\ &= \log_3 3+ 4 \log_3 3 \\ &= 5. \ _\square \end{align}\]
What is(are) the solution(s) of the quadratic equation \[\log 2x + \log(x-1) = \log\big(x^2+3\big) ?\] We have \[ \begin{align} \log 2x + \log(x-1) &= \log(x^2+3) \\ \log 2x(x-1) &= \log (x^2+3) \\ \Rightarrow 2x(x-1) &= x^2 +3 \\ x^2-2x-3 &= 0 \\ (x+1)(x-3) &= 0 \\ x &= -1, 3. \end{align} \] Since the logarithm functions \( \log(x-1)\) and \( \log 2x\) are defined over positive numbers, it must be true that \(x-1>0 \implies x>1\) and \(2x>0 \implies x>0.\) Thus, \(-1\) is can not be the value of \(x,\) implying that the value of \(x\) satisfying the given equation is \(x=3.\) \(_\square\)
What is the solution(s) of the quadratic equation \[ 2(\log x)^2 = 7\log x - 3 ?\] We have \[\begin{align} 2(\log x)^2 &= 7\log x - 3 \\ 2(\log x)^2 - 7\log x +3 &= 0 \\ (\log x -3)(2\log x -1) &= 0 \\ \Rightarrow \log x &= 3, \frac{1}{2} \\ x &= 1000, \sqrt{10}. \ _\square \end{align} \]
If the solutions of the quadratic equation \( x^{\log_3 x\,-\,2} = 27 \) are \(a\) and \(b,\) what is \( \log_{a} b + \log_{b} a?\) Taking logs with base 3 on both sides, we have \[ \begin{align} x^{\log_{3} x\,-\,2} &= 27 \\ \Rightarrow (\log_{3} x -2)\log_{3} x &= \log_{3} 27 \\ (\log_{3} x)^2-2\log_{3} x - 3 &= 0 \\ (\log_{3} x +1)(\log_{3} x - 3) &= 0 \\ \log_{3} x &= -1, 3. \end{align} \] Since \( \log_{a} b + \log_{b} a \) can be expressed as \(\frac{\log_{3} b}{\log_{3} a} + \frac{\log_{3} a}{\log_{3} b}\) using log with base 3, \[ \begin{align} \log_{a} b + \log_{b} a &= \frac{\log_{3} b}{\log_{3} a} + \frac{\log_{3} a}{\log_{3} b} \\ &= \frac{-1}{3} + \frac{3}{-1} \\ &= -\frac{10}{3}. \ _\square \end{align} \]
If the solutions of the equation \(\log_{2} x + a\log_{x} 8 = b \) are \(2\) and \(\frac{1}{8},\) what are \(a\) and \(b?\) We have \[ \begin{align} \log_{2} x + a\log_{x} 8 &= b \\ \log_{2} x + a\log_{x} 2^3 &= b \\ \log_{2} x + \frac{3a}{\log_{2} x} &= b \\ (\log_{2} x)^2 -b \log_{2} x + 3a &= 0. \qquad (1) \\ \end{align} \] Since the solutions of the equation \( {(\log_{2} x)}^2 -b \log_{2} x + 3a = 0 \) are \(2\) and \(\frac{1}{8} ,\) substituting \(2\) and \(\frac{1}{8}\) into \((1)\) gives \[ \begin{align} (\log_{2} 2)^2 - b \log_{2} 2 + 3a &= 0 \\ \Rightarrow 1-b+3a &= 0, \qquad (2)\\ (\log_{2} \frac{1}{8} )^2 - b \log_{2} \frac{1}{8} + 3a &= 0 \\ \Rightarrow 9+3b+3a &=0. \qquad (3) \end{align} \] Solving the simultaneous equations \((2)\) and \((3)\) gives \(a= -1\) and \(b = -2.\) \(_\square\)

The following logarithms are in an arithmetic progression:

\[\log_{2}4 + \log_{2}{16} + \log_{2}{64} + \cdots + x = 42.\]

If \(x\) can be expressed as \(\log_{2}a,\) find the value of \(a.\)

What are the solutions of the equation \[ \log_{x} xy \times \log_{y} xy + \log_{x}(x-y) \times \log_{y} (x-y) = 0?\] We have \[ \begin{align} \log_{x} xy \times \log_{y} xy + \log_{x}(x-y) \times \log_{y} (x-y) &= 0 \\ \frac{\log xy}{\log x} \times \frac{\log xy}{\log y} + \frac{\log(x-y)}{\log x} \times \frac{\log(x-y)}{\log y} &= 0 \\ \frac{(\log xy)^2 + (\log(x-y))^2}{\log x \cdot \log y} &= 0 \\ (\log xy)^2 + (\log(x-y))^2 &= 0 \\ \Rightarrow \log xy &= 0 \text{ and } \log(x-y)= 0. \\ \end{align} \] Since \(x\) and \(y\) are both positive, this implies that \[ \begin{align} xy &= 1 \text{ and } x-y=1 \\ \Rightarrow x&= \frac{\sqrt{5} +1}{2}, y=\frac{\sqrt{5}-1}{2}. \ _\square \end{align} \]

\[ \] Richter Scale:

Richter scale was developed by Charles Richter in 1935 to compare the intensities of earthquakes. The amount of energy released in an earthquake is very large, so a logarithmic scale avoids the use of large numbers.

The formula used for these calculations is

\[M= \log_{10}\left(\frac{I}{I_0}\right),\]

where \(M\) is the magnitude on the Richter scale, \(I\) is the intensity of the earthquake being measured, and \(I_0\) is the intensity of a reference earthquake.

Let's do a quick example to clarify how this works.

The 1906 San Francisco earthquake had a magnitude of 8.3 on the Richter scale. At the same time in South America there was an eathquake with magnitude 4.1 that caused only minor damage. How many times more intense was the San Francisco earthquake than the South American one?

Because the magnitude is a base-10 log, the Richter number is actually the exponent that 10 is raised to in order to calculate the intensity of the earthquake. Thus, the difference in magnitudes of the earthquakes can be calculated as follows:

\[M=\log_{10}\left(\frac{10^{8.3}}{10^{4.1}}\right)=4.2.\]

So, to answer the question, the San Francisco earthquake is more intense than the South American one by about \(10^M \approx 15848.93192\) times!

Note that you can just subtract 4.1 from 8.3 and get the same result. But if your math teachers are like mine, they will want you to use logarithms, and this is how it is done. The reason that subtracting the magnitudes works is because of the exponent rule for dividing exponents with the same base.

\[ \] Decibel Scale:

One decibel is one tenth of one bel, named in honor of Alexander Graham Bell. The bel is rarely used without the deci- prefix, deci- meaning one tenth. The decibel scale is used to calculate the difference in intensity between two sounds:

\[L=10\log_{10}\left(\frac{I}{I_0}\right),\]

where \( L\) is the loudness of the sound measured in decibels, \(I\) is the intensity of the sound being measured, and \(I_0\) is the intensity of the sound at the threshold of hearing which is equal to zero decibels.

\[ \] \(\text{pH}\) Scale:

The \(\text{pH}\) scale was invented in 1910 by Dr. Soren Sorenson, Head of Laboratory at Carlsberg Beer Company. The "H" in \(\text{pH}\) stands for hydrogen and the meaning of the "p" in \(\text{pH}\), although disputed, is generally considered to mean the power of hydrogen. This scale is used to measure the acidity or alkalinity of water or water soluble substances including, but definitely not limited to, soil or rainwater. The \(\text{pH}\) scale ranges from 1 to 14, where seven is a neutral point. Values below 7 indicate acidity with 1 being the most acidic. Values above 7 indicate alkalinity with14 being the most alkaline:

\[\text{pH}=-\log_{10}\ce{[H+]},\]

where \(\text{pH}\) is the \(\text{pH}\) number between \(1\) and \(14\) and \(\ce{[H+]}\) is the concentration of hydrogen ions.

  • Solving Logarithmic Equations
  • Graphs of Logarithmic Functions
  • Solving Logarithmic Inequalities

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Logarithms is another way of writing exponents. We know that 2 5 = 32. But if we are asked to find what number replaces the question mark in 2 ? = 32, then by trial and error, we can simply find that the answer is 5. But what if we are asked to find the question mark in 2 ? = 30? Is there any number such that 2 raised to it gives 30? No, then how to solve it? The solution is logarithms (or logs).

Let us learn more about logarithms along with their properties with examples.

What is Logarithm?

Logarithm is nothing but another way of expressing exponents and can be used to solve problems that cannot be solved using the concept of exponents only. Understanding logs is not so difficult. To understand logarithms, it is sufficient to know that a logarithmic equation is just another way of writing an exponential equation .

Logarithm and exponent are inverse forms of each other. One can understand this from the section below. Initially, a mathematician named John Napier introduced logarithms for making calculations in a simple way and this concept is quickly adopted by other scientists, engineers, etc.

Logarithms or logs

Here is the mathematical definition of logs.

Logs Definition

A logarithm is defined using an exponent .

  • b x = a ⇔ log b a = x

Here, "log" stands for logarithm. The right side part of the arrow is read to be "Logarithm of a to the base b is equal to x".

A very simple way to remember this is "base stays as the base in both forms" and "base doesn't stay with the exponent in log form". Notice that 'b' is the base both on the left and right sides of the implies symbol and in the log form see that the base b and the exponent x don't stay on the same side of the equation.

  • a and b are two positive real numbers .
  • x is a real number.
  • a, which is inside the log is called the "argument".
  • b, which is at the bottom of the log is called the "base".

logs definition or logarithm definition

The above equation has two things to understand (from the symbol ⇔):

  • b x = a ⇒ log b a = x. This is called " exponential to log form "
  • log b a = x ⇒ b x = a . This is called " log to exponential form "

Here is a table to understand the conversions from one form to the other form.

Natural Log and Common Log

Observe the last two rows of the above table. They have log e and log 10 . These two logs have specific importance and specific names in logarithms.

  • log e is called the natural log
  • log 10 is called the common log

Let us study more about each of these.

Natural Logarithm

Natural logarithm is nothing but log with base e . That is, a natural log means log e . But it is not usually represented as log e . Instead, it is represented as ln. i.e.,

  • e x = 2 ⇒ log e 2 = x (or) ln 2 = x.
  • e x = 7 ⇒ log e 7 = x (or) ln 7 = x.

Common Logarithm

Common logarithm is nothing but log with base 10. That is, a common log means log 10 . But usually, writing "log" is sufficient instead of writing log 10 . i.e.,

  • log 10 = log

i.e., if there is no base for a log it means that its log 10 . In other words, it is a common logarithm.

  • 10 2 = 100 ⇒ log 10 100 = 2 (or) log 100 = 2
  • 10 -2 = 0.01 ⇒ log 10 0.01 = -2 (or) log 0.01 = -2

Observe that we have not written 10 as the base in these examples, because that's obvious.

Rules of Logs

The rules of logs are used to simplify a logarithm, expand a logarithm, or compress a group of logarithms into a single logarithm. Here are the rules (or) properties of logs. If you want to see how all these rules are derived, click here .

rules of logs

Let us see each of these rules one by one here.

The value of log 1 irrespective of the base is 0. Because from the properties of exponents, we know that, a 0 = 1, for any 'a'. Converting this into log form, log a 1 = 0, for any 'a'. Obviously, when a = 10, log 10 1 = 0 (or) simply log 1 = 0.

When we extend this to the natural logarithm, we have, since e 0 = 1 ⇒ ln 1 = 0.

Since a 1 = a, for any 'a', converting this equation into log form, log a a = 1. Thus, the logarithm of any number to the same base is always 1. For example:

  • log 2 2 = 1
  • log 3 3 = 1

Product Rule of Log

The logarithm of a product of two numbers is the sum of the logarithms of the individual numbers, i.e.,

  • log a mn = log a m + log a n

Note that the bases of all logs must be the same here. This resembles/is derived from the product rule of exponents : x m ⋅ x n = x m+n .

  • log 6 = log (3 x 2) = log 3 + log 2
  • log (5x) = log 5 + log x

Quotient Rule of Log

The logarithm of a quotient of two numbers is the difference between the logarithms of the individual numbers, i.e.,

  • log a (m/n) = log a m - log a n

Note that the bases of all logs must be the same here as well. This resembles/is derived from the quotient rule of exponents: x m / x n = x m-n .

  • log 4 = log (8/2) = log 8 - log 2
  • log (x/2) = log x - log 2

Power Rule of Log

The exponent of the argument of a logarithm can be brought in front of the logarithm, i.e.,

  • log a m n = n log a m

Here, the bases must be the same on both sides. This resembles/is derived from the power of power rule of exponents: (x m ) n = x mn .

Change of Base Rule

The base of a logarithm can be changed using this property. It says:

  • log b a = (log꜀ a) / (log꜀ b)

Another way of writing this rule is log b a · log꜀ b = log꜀ a.

Using this property, we can change the base to any other number. Hence we can change the base to 10 as well. Then we get: log b a = (log a) / (log b). Thus:

  • log 2 3 = (log 3) / (log 2)
  • log 3 2 = (log 2) / (log 3)

Equality Rule of Logarithms

This rule is used while solving the equations involving logarithms. i.e.,

  • log b a = log b c ⇒ a = c

It is a kind of canceling log from both sides.

Number Raised to Log Property

When a number is raised to log whose base is same as the number, then the result is just the argument of the logarithm. i.e.,

  • a log a x = x

Here are some examples of this property.

  • 2 log 2 5 = 5
  • 10 log 6 = 6

Negative Log Property

The negative logs are of the form −log b a. We can calculate this using the power rule of logarithms.

−log b a = log b a -1 = log b (1/a)

  • −log b a = log b (1/a)

i.e., To convert a negative log into a positive log, we can just take the reciprocal of the argument. Also, to convert a negative log into a positive log, we can take the reciprocal of the base, i.e.,

  • −log b a = log 1/b a

How to Condense/Expand Logarithms?

We can either compress a group of logs into a single log or expand a single log into a group of logs using the above rules of logs. But the important rules that we use in this process are:

  • log a mn = log a m + log a n (Product rule of logarithms)
  • log a m/n = log a m - log a n (Quotient rule of logarithms)
  • log a m n = n log a m (Power rule of logarithms)

Expanding Logarithms

Let us expand the logarithm log (3x 2 y 3 ).

log (3x 2 y 3 ) = log (3) + log (x 2 ) + log (y 3 ) (By product rule) = log 3 + 2 log x + 3 log y (By power rule)

Condensing Logarithms

Let us just take the above sum of logarithms and compress it. We should get log (3x 2 y 3 ) back.

log 3 + 2 log x + 3 log y = log (3) + log (x 2 ) + log (y 3 ) (By power rule) = log (3x 2 y 3 ) (By product rule)

Important Notes on Logarithms:

  • The logarithm of 0 is NOT defined as one number raised to another number never gives 0 as the result.
  • An exponential equation is converted into a logarithmic equation and vice versa using b x = a ⇔ log b a = x.
  • A common log is a logarithm with base 10, i.e., log 10 = log.
  • A natural log is a logarithm with base e, i.e., log e = ln.
  • Logarithms are used to do the most difficult calculations of multiplication and division .

☛ Related Topics:

  • Common Log Calculator
  • Natural Log Calculator

Solved Examples on Logarithms

Example 1: Find the value of log 5 (1/25).

We will simplify the given expression using the rules of logarithms.

log 5 (1/25) = log 5 (1/5 2 ) = log 5 (5 -2 ) = -2 log 5 5 = -2(1) = -2

Answer: log 5 (1/25) = -2.

Example 2: Expand the logarithm log 6 \(\left(\frac{6 \mathrm{~m}^{3}}{\sqrt{\mathrm{n}}}\right)\).

By using the rules of logs,

log 6 \(\left(\frac{6 \mathrm{~m}^{3}}{\sqrt{\mathrm{n}}}\right)\) = log 6 (6m 3 ) - log 6 √n = log 6 6 + log 6 m 3 - log 6 n 1/2 = 1 + 3 log 6 m - (1/2) log 6 n

Answer: The expanded form is 1 + 3 log 6 m - (1/2) log 6 n.

Example 3: Compress the following as a single logarithm: 5 log 3 x + 2 log 3 (4x) - log 3 (8x 5 ).

By the properties of logarithms,

5 log 3 x + 2 log 3 (4x) - log 3 (8x 5 ) = log 3 x 5 + log 3 (4x) 2 - log 3 (8x 5 ) = log 3 x 5 + log 3 (16x 2 ) - log 3 (8x 5 ) = log 3 (x 5 · 16x 2 ) - log 3 (8x 5 ) = log 3 (16x 7 ) - log 3 (8x 5 ) = log 3 (16x 7 / 8x 5 ) = log 3 (2x 2 )

Answer: The compressed logarithmic form is log 3 (2x 2 )

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Practice Questions on Logs in Math

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FAQs on Logarithms

What are logs in math.

Logs are the other way of writing exponent. The formula for conversion between exponential and log forms is: b x = a ⇔ log b a = x. Logarithms are very useful in solving equations involving exponents.

What are the Values of Logarithms log 0, log 1, log 2, log 3, log 4, log 5, log 10, log 100, and log inf?

Here are the values of the given logs:

  • log 0 is not defined for any base because a number raised to any number doesn't result in 0.
  • log 1 = 0 as 10 0 = 1
  • log 2 ≈ 0.3010 (using calculator)
  • log 3 ≈ 0.4771 (using calculator)
  • log 4 ≈ 0.6021 (using calculator)
  • log 5 ≈ 0.6990 (using calculator)
  • log 10 = 1 as 10 1 = 10
  • log 100 = 2 as 10 2 = 100

What is ln in Math?

Ln in math is used to represent the natural logarithms. i.e., ln = "log with base e". For example, e 2 = x ⇔ ln x = 2.

What are 3 Types of Logarithms?

There are basically three types of logarithms:

  • Common logarithm which is written as log without base. For example: log 2
  • Natural logarithm which is written as "ln" (means log e ). For example: ln 2
  • Logarithm with any other base (no specific name). For example: log 3 2.

What are the Values of Logarithms ln e, ln 1, and ln of 0?

Here the values of the given natural logs.

  • ln e = 1 as e 1 = e
  • ln 1 = 0 as e 0 = 1
  • ln 0 is NOT defined

What are Important Logarithmic Properties?

The important logarithmic properties are:

  • Product rule: log a mn = log a m + log a n
  • Quotient rule: log a m/n = log a m - log a n
  • Power rule: log a m n = n log a m

How do You Calculate Logs?

We can calculate logs using the properties of logarithms . i.e., using the rules of logs we can either compress a set of logarithms into one or expand one logarithm as many. We also use log table and antilog table in calculations.

What is the Derivative of ln x and log x?

Here are the derivatives:

  • The derivative of ln x is d/dx (ln x) = 1/x.
  • The derivative of log x is d/dx(log a x) (or) (log a x)' = 1/(x ln a).

What is the Integral of ln x and log x?

Here are the given integrals:

  • The integral of ln x is ∫ln x dx = x ln x - x + C.
  • The integral of log x is ∫log x dx = x log x - x/ln 10 + C.

What is log ln e?

We know that ln e = 1 and log 1 = 0. Using these two facts, log ln e = log 1 = 0.

Is Log Square x Same as 2 Log x?

No, log square x is NOT the same as 2 log x. Observe the following.

  • log x square = log x 2 = 2 log x (using power rule)
  • log square x = log 2 x = (log x) 2 = (log x) (log x) and this can't be simplified further using any rule.

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  • Solving Logarithmic Functions – Explanation & Examples

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Properties of logarithmic functions

Comparison of exponential function and logarithmic function, practice questions, solving logarithmic functions – explanation & examples.

Solving Log Function Title

Logarithms and exponents are two topics in mathematics that are closely related. Therefore it is useful we take a brief review of exponents.

An exponent is a form of writing the repeated multiplication of a number by itself. An exponential function is of the form f (x) = b y , where b > 0 < x and b ≠ 1. The quantity x is the number, b is the base, and y is the exponent or power.

For example , 32 = 2 × 2 × 2 × 2 × 2 = 2 2 .

Solving Log Function Exp and Log

On the other hand, the logarithmic function is defined as the inverse function of exponentiation. Consider again the exponential function f(x) = b y , where b > 0 < x and b ≠ 1. We can represent this function in logarithmic form as:

y = log b x

Then the logarithmic function is given by;

f(x) = log b x = y, where b is the base, y is the exponent, and x is the argument.

The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in mathematics because they enable us to perform calculations with very large numbers.

How to Solve Logarithmic Functions?

To solve the logarithmic functions, it is important to use exponential functions in the given expression. The natural log or ln is the inverse of e . That means one can undo the other one i.e.

ln (e x ) = x

To solve an equation with logarithm(s), it is important to know their properties.

Properties of logarithmic functions are simply the rules for simplifying logarithms when the inputs are in the form of division, multiplication, or exponents of logarithmic values.

Some of the properties are listed below.

  • Product rule

The product rule of logarithm states the logarithm of the product of two numbers having a common base is equal to the sum of individual logarithms.

⟹ log a  (p q) = log a  p + log a  q.

  • Quotient rule

The quotient rule of logarithms states that the logarithm of the two numbers’ ratio with the same bases is equal to the difference of each logarithm.

⟹ log a  (p/q) = log a  p – log a q

The power rule of logarithm states that the logarithm of a number with a rational exponent is equal to the product of the exponent and its logarithm.

⟹ log a  (p q ) = q log a p

  • Change of Base rule

⟹ log a p = log x p ⋅ log a x

⟹ log q p = log x p / log x q

  • Zero Exponent Rule

Solving Log Function Properties

Other properties of logarithmic functions include:

  • The bases of an exponential function and its equivalent logarithmic function are equal.
  • The logarithms of a positive number to the base of the same number are equal to 1.

log a  a = 1

  • Logarithms of 1 to any base are 0.

log a  1 = 0

  • Log a 0 is undefined
  • Logarithms of negative numbers are undefined.
  • The base of logarithms can never be negative or 1.
  • A logarithmic function with base 10is called a common logarithm. Always assume a base of 10 when solving with logarithmic functions without a small subscript for the base.

Whenever you see logarithms in the equation, you always think of how to undo the logarithm to solve the equation. For that, you use an exponential function . Both of these functions are interchangeable.

The following table tells the way of writing and interchanging the exponential functions and logarithmic functions . The third column tells about how to read both the logarithmic functions.

Let’s use these properties to solve a couple of problems involving logarithmic functions.

Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function.

Given 7 2 = 64.

Here, the base = 7, exponent = 2 and the argument = 49. Therefore, 7 2 = 64 in logarithmic function is;

⟹ log 7 49 = 2

Write the logarithmic equivalent of 5 3 = 125.

exponent = 3;

and argument = 125

5 3 = 125 ⟹ log 5 125 =3

Solve for x in log  3  x = 2

log  3  x = 2 3 2  = x ⟹ x = 9

If 2 log x = 4 log 3, then find the value of ‘x’.

2 log x = 4 log 3

Divide each side by 2.

log x = (4 log 3) / 2

log x = 2 log 3

log x = log 3 2

log x = log 9

Find the logarithm of 1024 to the base 2.

1024 = 2 10

log 2 1024 = 10

Find the value of x in log 2 ( x ) = 4

Rewrite the logarithmic function log 2 ( x ) = 4 to exponential form.

Solve for x in the following logarithmic function log 2 (x – 1) = 5.

Solution Rewrite the logarithm in exponential form as;

log 2 (x – 1) = 5 ⟹ x – 1 = 2 5

Now, solve for x in the algebraic equation. ⟹ x – 1 = 32 x = 33

Find the value of x in log x 900 = 2.

Write the logarithm in exponential form as;

Find the square root of both sides of the equation to get;

x = -30 and 30

But since, the base of logarithms can never be negative or 1, therefore, the correct answer is 30.

Solve for x given, log x = log 2 + log 5

Using the product rule Log b  (m n) = log b  m + log b  n we get;

⟹ log 2 + log 5 = log (2 * 5) = Log   (10).

Therefore, x = 10.

Solve log  x  (4x – 3) = 2

Rewrite the logarithm in exponential form to get;

x 2  = 4x – 3

Now, solve the quadratic equation. x 2  = 4x – 3 x 2  – 4x + 3 = 0 (x -1) (x – 3) = 0

Since the base of a logarithm can never be 1, then the only solution is 3.

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Introduction to Logarithms

In its simplest form, a logarithm answers the question:

How many of one number multiply together to make another number?

Example: How many 2 s multiply together to make 8 ?

Answer: 2 × 2 × 2 = 8 , so we had to multiply 3 of the 2 s to get 8

So the logarithm is 3

How to Write it

We write it like this:

log 2 (8) = 3

 So these two things are the same:

The number we multiply is called the "base", so we can say:

  • "the logarithm of 8 with base 2 is 3"
  • or "log base 2 of 8 is 3"
  • or "the base-2 log of 8 is 3"

Notice we are dealing with three numbers:

  • the base : the number we are multiplying (a "2" in the example above)
  • how often to use it in a multiplication (3 times, which is the logarithm )
  • The number we want to get (an "8")

More Examples

Example: what is log 5 (625) ... .

We are asking "how many 5s need to be multiplied together to get 625?"

5 × 5 × 5 × 5 = 625 , so we need 4 of the 5s

Answer: log 5 (625) = 4

Example: What is log 2 (64) ... ?

We are asking "how many 2s need to be multiplied together to get 64?"

2 × 2 × 2 × 2 × 2 × 2 = 64 , so we need 6 of the 2s

Answer: log 2 (64) = 6

Exponents and Logarithms are related, let's find out how ...

So a logarithm answers a question like this:

In this way:

The logarithm tells us what the exponent is!

In that example the "base" is 2 and the "exponent" is 3:

So the logarithm answers the question:

What exponent do we need (for one number to become another number) ?

The general case is:

Example: What is log 10 (100) ... ?

So an exponent of 2 is needed to make 10 into 100, and:

log 10 (100) = 2

Example: What is log 3 (81) ... ?

So an exponent of 4 is needed to make 3 into 81, and:

log 3 (81) = 4

Common Logarithms: Base 10

Sometimes a logarithm is written without a base, like this:

This usually means that the base is really 10 .

It is called a "common logarithm". Engineers love to use it.

On a calculator it is the "log" button.

It is how many times we need to use 10 in a multiplication, to get our desired number.

Example: log(1000) = log 10 (1000) = 3

Natural Logarithms: Base "e"

Another base that is often used is e (Euler's Number) which is about 2.71828.

This is called a "natural logarithm". Mathematicians use this one a lot.

On a calculator it is the "ln" button.

It is how many times we need to use "e" in a multiplication, to get our desired number.

Example: ln(7.389) = log e ( 7.389 ) ≈ 2

Because 2.71828 2 ≈ 7.389

But Sometimes There Is Confusion ... !

Mathematicians may use "log" (instead of "ln") to mean the natural logarithm. This can lead to confusion:

So, be careful when you read "log" that you know what base they mean!

Logarithms Can Have Decimals

All of our examples have used whole number logarithms (like 2 or 3), but logarithms can have decimal values like 2.5, or 6.081, etc.

Example: what is log 10 (26) ... ?

The logarithm is saying that 10 1.41497... = 26 (10 with an exponent of 1.41497... equals 26)

Read Logarithms Can Have Decimals to find out more.

Negative Logarithms

A negative logarithm means how many times to divide by the number.

We can have just one divide:

Example: What is log 8 (0.125) ... ?

Well, 1 ÷ 8 = 0.125 ,

So log 8 (0.125) = −1

Or many divides:

Example: What is log 5 (0.008) ... ?

1 ÷ 5 ÷ 5 ÷ 5 = 5 -3 ,

So log 5 (0.008) = −3

It All Makes Sense

Multiplying and Dividing are all part of the same simple pattern.

Let us look at some Base-10 logarithms as an example:

Looking at that table, see how positive, zero or negative logarithms are really part of the same (fairly simple) pattern.

"Logarithm" is a word made up by Scottish mathematician John Napier (1550-1617), from the Greek word logos meaning "proportion, ratio or word" and arithmos meaning "number", ... which together makes "ratio-number" !

Common and Natural Logarithm

In these lessons, we will learn common logarithms and natural logarithms and how to solve problems using common log and natural log.

Related Pages Logarithmic Functions Properties Or Rules Of Logarithms Rules Of Exponents Logarithm Rules

The following diagrams gives the definition of Logarithm, Common Log, and Natural Log. Scroll down the page for more examples and solutions.

Definition of Logarithm

Common Logarithms

Logarithms to base 10 are called common logarithms . We often write “log 10 ” as “log” or “lg”. Common logarithms can be evaluated using a scientific calculator.

Recall that by the definition of logarithm. log Y = X ↔ Y = 10 X

Natural Logarithms

Besides base 10, another important base is e. Log to base e are called natural logarithms . “log e ” are often abbreviated as “ln”. Natural logarithms can also be evaluated using a scientific calculator.

By definition ln Y = X ↔ Y = e X

Using a calculator, we can use common and natural logarithms to solve equations of the form a x = b, especially when b cannot be expressed as a n .

Example: Solve the equations a) 6 x + 2 = 21 b) e 2x = 9

Solution: a) 6 x + 2 = 21 log 6 x + 2 = log 21 (x + 2) log 6 = log 21

b) e 3x = 9 ln e 3x = ln 9 3x ln e = ln 9 3x = ln 9

Example: Express 3 x (2 2x ) = 7(5 x ) in the form a x = b. Hence, find x.

Solution: Since 3 x (2 2x ) = 3 x (2 2 ) x = (3 × 4) x = 12 x the equation becomes

12 x = 7(5 x )

Common And Natural Logarithms

We can use many bases for a logarithm, but the bases most typically used are the bases of the common logarithm and the natural logarithm. The common logarithm has base 10, and is represented on the calculator as log(x). The natural logarithm has base e, a famous irrational number, and is represented on the calculator by ln(x). The natural and common logarithm can be found throughout Algebra and Calculus.

Defines common log, log x, and natural log, ln x, and works through examples and problems using a calculator.

Example: Write the following logarithms in exponential form. Evaluate if possible.

Properties Of Logarithms

The logarithm of a Product: log b MN = log b M + log b N

The logarithm of a Quotient: log b M/N = log b M - log b N

The logarithm of a number raised to a power: log b M P = P log b M

How to use the properties of logarithms to condense and solve logarithms? How to use the properties of logarithms to expand logarithms?

Common And Natural Logs

Examples: Solve without a calculator: log 3 3 log 1 log 16 2 ln e 3 Solve with a calculator: log 3 log 32 ln √5 ln 7.3

How to solve logarithmic equations?

The first example is with common logs and the second example is natural logs. It is good to remember the properties of logarithms also can be applied to natural logs.

Examples: Solve, round to four decimal places.

  • log x = log2x 2 - 2
  • ln x + ln (x + 1) = 5

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Logarithm Questions

Logarithm questions with answers are provided for students to solve them and understand the concept elaborately. These questions are based on the logarithm chapter of Class 9, 10 and 11 syllabi. Practising these problems will not only help students to score good marks in academic exams but also participate in competitive exams conducted at the state or national level, such as Maths Olympiad.

The logarithmic function is an inverse of the exponential function. It is defined as:

y=log a x, if and only if x=a y ; for x>0, a>0, and a≠1.

Natural logarithmic function: The log function with base e is called natural logarithmic function and is denoted by log e .

f(x) = log e x

The questions of logarithm could be solved based on the properties, given below:

Also, read:

  • Logarithm Table

Questions on Logarithm with Solutions

1. Express 5 3 = 125 in logarithm form.

As we know,

a b = c ⇒ log a c=b

Log 5 125 = 3

2. Express log 10 1 = 0 in exponential form.

Given, log 10 1 = 0

By the rule, we know;

log a c=b ⇒ a b = c

3. Find the log of 32 to the base 4.

Solution: log 4 32 = x

(2 2 ) x = 2x2x2x2x2

log 4 32 =5/2

4. Find x if log 5 (x-7)=1.

Solution: Given,

log 5 (x-7)=1

Using logarithm rules, we can write;

5. If log a m=n, express a n-1 in terms of a and m.

6. Solve for x if log(x-1)+log(x+1)=log 2 1

Solution: log(x-1)+log(x+1)=log 2 1

log(x-1)+log(x+1)=0

log[(x-1)(x+1)]=0

Since, log 1 = 0

(x-1)(x+1) = 1

Since, log of negative number is not defined.

Therefore, x=√2

7. Express log(75/16)-2log(5/9)+log(32/243) in terms of log 2 and log 3.

Solution: log(75/16)-2log(5/9)+log(32/243)

Since, nlog a m=log a m n

⇒log(75/16)-log(5/9) 2 +log(32/243)

⇒log(75/16)-log(25/81)+log(32/243)

Since, log a m-log a n=log a (m/n)

⇒log[(75/16)÷(25/81)]+log(32/243)

⇒log[(75/16)×(81/25)]+log(32/243)

⇒log(243/16)+log(32/243)

Since, log a m+log a n=log a mn

⇒log(32/16)

8. Express 2logx+3logy=log a in logarithm free form.

Solution: 2logx+3logy=log a

Video Lesson

Logarithmic equations.

logarithm problem solving examples

9. Prove that: 2log(15/18)-log(25/162)+log(4/9)=log2

Solution: 2log(15/18)-log(25/162)+log(4/9)=log2

Taking L.H.S.:

⇒2log(15/18)-log(25/162)+log(4/9)

⇒log(15/18) 2 -log(25/162)+log(4/9)

⇒log(225/324)-log(25/162)+log(4/9)

⇒log[(225/324)(4/9)]-log(25/162)

⇒log[(225/324)(4/9)]/(25/162)

⇒log(72/36)

⇒log2 (R.H.S)

10. Express log 10 (2) + 1 in the form of log 10 x.

Solution: log 10 (2)+1

11. Find the value of x, if log 10 (x-10)=1.

Solution: Given, log 10 (x-10)=1.

log 10 (x-10) = log 10 10

12. Find the value of x, if log(x+5)+log(x-5)=4log2+2log3

log(x+5)+log(x-5)=4log2+2log3

log(x 2 -25) = log2 4 +log3 2

log(x 2 -25) = log16+log9

log(x 2 -25)=log(16×9)

log(x 2 -25)=log144

x 2 -25=144

13. Solve for x, if (log 225/log15) = log x

Solution: log x = (log 225/log15)

log x = log 15 2 /log 15

log x = 2log 15/log 15

x=10×10

Practice Questions

  • If log x = m+n and log y=m-n, express the value of log 10x/y 2 in terms of m and n.
  • Express 3 -2 =1/9 in logarithmic form.
  • Express log 10 0.01=-2 in exponential form.
  • Find the logarithm of 1/81 to the base 27.
  • Find x if log 7 (2x 2 -1)=2.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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Mathematics LibreTexts

5.6: Application Problems with Exponential and Logarithmic Functions

  • Last updated
  • Save as PDF
  • Page ID 38600

  • Rupinder Sekhon and Roberta Bloom
  • De Anza College

Learning Objectives

In this section, you will:

  • review strategies for solving equations arising from exponential formulas
  • solve application problems involving exponential functions and logarithmic functions

STRATEGIES FOR SOLVING EQUATIONS THAT CONTAIN EXPONENTS

When solving application problems that involve exponential and logarithmic functions, we need to pay close attention to the position of the variable in the equation to determine the proper way solve the equation we investigate solving equations that contain exponents.

Suppose we have an equation in the form : value = coefficient(base) exponent

We consider four strategies for solving the equation:

STRATEGY A : If the coefficient, base, and exponent are all known, we only need to evaluate the expression for coefficient(base) exponent to evaluate its value.

STRATEGY B: If the variable is the coefficient, evaluate the expression for (base) exponent . Then it becomes a linear equation which we solve by dividing to isolate the variable.

STRATEGY C : If the variable is in the exponent, use logarithms to solve the equation.

STRATEGY D: If the variable is not in the exponent, but is in the base, use roots to solve the equation.

Below we examine each strategy with one or two examples of its use.

Example \(\PageIndex{1}\)

Suppose that a stock’s price is rising at the rate of 7% per year, and that it continues to increase at this rate. If the value of one share of this stock is $43 now, find the value of one share of this stock three years from now.

The problem tells us that \(a\) = 43 and \(r\) = 0.07, so \(b = 1+ r = 1+ 0.07 = 1.07\)

Therefore, function is \(y = 43(1.07)^t\).

In this case we know that \(t\) = 3 years, and we need to evaluate \(y\) when \(t\) = 3.

At the end of 3 years, the value of this one share of this stock will be

\[y=43(1.07)^{3}=\$ 52.68 \nonumber \]

Example \(\PageIndex{2}\)

The value of a new car depreciates (decreases) after it is purchased. Suppose that the value of the car depreciates according to an exponential decay model. Suppose that the value of the car is $12000 at the end of 5 years and that its value has been decreasing at the rate of 9% per year. Find the value of the car when it was new.

The function is \(y = a(0.91)^t\)

In this case we know that when \(t\) = 5, then \(y\) = 12000; substituting these values gives

\[12000 = a(0.91)^5 \nonumber \]

We need to solve for the initial value a, the purchase price of the car when new.

First evaluate (0.91) 5 ; then solve the resulting linear equation to find \(a\).

\[ 1200 = a(0.624) \nonumber \]

\(a=\frac{12000}{0.624} = \$ 19,230.77\); The car's value was $19,230.77 when it was new.

Example \(\PageIndex{3}\)

A national park has a population of 5000 deer in the year 2016. Conservationists are concerned because the deer population is decreasing at the rate of 7% per year. If the population continues to decrease at this rate, how long will it take until the population is only 3000 deer?

\(r\) = -0.07 and \(b = 1+r = 1+(-0.07) = 0.93\) and the initial population is \(a\) = 5000

The exponential decay function is \(y = 5000(0.93)^t\)

To find when the population will be 3000, substitute \(y\) = 3000

\[ 3000 = 5000(0.93)^t \nonumber \]

Next, divide both sides by 5000 to isolate the exponential expression

\[\begin{array}{l} \frac{3000}{5000}=\frac{5000}{5000}(0.93)^{2} \\ 0.6=0.93^{t} \end{array} \nonumber \]

Rewrite the equation in logarithmic form; then use the change of base formula to evaluate.

\[t=\log _{0.93}(0.6) \nonumber \]

\(t = \frac{\ln(0.6)}{\ln(0.93)}=7.039\) years; After 7.039 years, there are 3000 deer.

Note: In Example \(\PageIndex{3}\), we needed to state the answer to several decimal places of precision to remain accurate. Evaluating the original function using a rounded value of \(t\) = 7 years gives a value that is close to 3000, but not exactly 3000.

\[y=5000(0.93)^{7}=3008.5 \text { deer } \nonumber \]

However using \(t\) = 7.039 years produces a value of 3000 for the population of deer

\[ y=5000(0.93)^{7.039}=3000.0016 \approx 3000 \text { deer } \nonumber \]

Example \(\PageIndex{4}\)

A video posted on YouTube initially had 80 views as soon as it was posted. The total number of views to date has been increasing exponentially according to the exponential growth function \(y = 80e^{0.2t}\), where \(t\) represents time measured in days since the video was posted. How many days does it take until 2500 people have viewed this video?

Let \(y\) be the total number of views \(t\) days after the video is initially posted. We are given that the exponential growth function is \(y = 80e^{0.2t}\) and we want to find the value of \(t\) for which \(y\) = 2500. Substitute \(y\) = 2500 into the equation and use natural log to solve for \(t\).

\[2500 = 80e^{0.12t} \nonumber \]

Divide both sides by the coefficient, 80, to isolate the exponential expression.

\[\begin{array}{c} \frac{2500}{80}=\frac{80}{80} e^{0.12 t} \\ 31.25=e^{0.12 t} \end{array} \nonumber \]

Rewrite the equation in logarithmic form

\[ 0.12t = \ln(31.25) \nonumber \]

Divide both sides by 0.04 to isolate \(t\); then use your calculator and its natural log function to evaluate the expression and solve for \(t\).

\[\begin{array}{l} \mathrm{t}=\frac{\ln (31.25)}{0.12} \\ \mathrm{t}=\frac{3.442}{0.12} \\ \mathrm{t} \approx 28.7 \text { days } \end{array} \nonumber \]

This video will have 2500 total views approximately 28.7 days after it was posted.

STRATEGY D: If the variable is not in the exponent, but is in the base, we use roots to solve the equation. It is important to remember that we only use logarithms when the variable is in the exponent.

Example \(\PageIndex{5}\)

A statistician creates a website to analyze sports statistics. His business plan states that his goal is to accumulate 50,000 followers by the end of 2 years (24 months from now). He hopes that if he achieves this goal his site will be purchased by a sports news outlet. The initial user base of people signed up as a result of pre-launch advertising is 400 people. Find the monthly growth rate needed if the user base is to accumulate to 50,000 users at the end of 24 months.

Let \(y\) be the total user base \(t\) months after the site is launched.

The growth function for this site is \(y = 400(1+r)^t\);

We don’t know the growth rate \(r\). We do know that when \(t\) = 24 months, then \(y\) = 50000.

Substitute the values of \(y\) and \(t\); then we need to solve for \(r\).

\[5000 = 400(1+r)^{24} \nonumber \]

Divide both sides by 400 to isolate (1+r) 24 on one side of the equation

\[\begin{array}{l} \frac{50000}{400}=\frac{400}{400}(1+r)^{24} \\ 125=(1+r)^{24} \end{array} \nonumber \]

Because the variable in this equation is in the base, we use roots:

\[\begin{array}{l} \sqrt[24]{125}=1+r \\ 125^{1 / 24}=1+r \\ 1.2228 \approx 1+r \\ 0.2228 \approx r \end{array} \nonumber \]

The website’s user base needs to increase at the rate of 22.28% per month in order to accumulate 50,000 users by the end of 24 months.

Example \(\PageIndex{6}\)

A fact sheet on caffeine dependence from Johns Hopkins Medical Center states that the half life of caffeine in the body is between 4 and 6 hours. Assuming that the typical half life of caffeine in the body is 5 hours for the average person and that a typical cup of coffee has 120 mg of caffeine.

  • Write the decay function.
  • Find the hourly rate at which caffeine leaves the body.
  • How long does it take until only 20 mg of caffiene is still in the body? www.hopkinsmedicine.org/psyc...fact_sheet.pdf

a. Let \(y\) be the total amount of caffeine in the body \(t\) hours after drinking the coffee.

Exponential decay function \(y = ab^t\) models this situation.

The initial amount of caffeine is \(a\) = 120.

We don’t know \(b\) or \(r\), but we know that the half- life of caffeine in the body is 5 hours. This tells us that when \(t\) = 5, then there is half the initial amount of caffeine remaining in the body.

\[\begin{array}{l} y=120 b^{t} \\ \frac{1}{2}(120)=120 b^{5} \\ 60=120 b^{5} \end{array} \nonumber \]

Divide both sides by 120 to isolate the expression \(b^5\) that contains the variable.

\[\begin{array}{l} \frac{60}{120}=\frac{120}{120} \mathrm{b}^{5} \\ 0.5=\mathrm{b}^{5} \end{array} \nonumber \]

The variable is in the base and the exponent is a number. Use roots to solve for \(b\):

\[\begin{array}{l} \sqrt[5]{0.5}=\mathrm{b} \\ 0.5^{1 / 5}=\mathrm{b} \\ 0.87=\mathrm{b} \end{array} \nonumber \]

We can now write the decay function for the amount of caffeine (in mg.) remaining in the body \(t\) hours after drinking a cup of coffee with 120 mg of caffeine

\[y=f(t)=120(0.87)^{t} \nonumber \]

b. Use \(b = 1 + r\) to find the decay rate \(r\). Because \(b = 0.87 < 1\) and the amount of caffeine in the body is decreasing over time, the value of \(r\) will be negative.

\[\begin{array}{l} 0.87=1+r \\ r=-0.13 \end{array} \nonumber \]

The decay rate is 13%; the amount of caffeine in the body decreases by 13% per hour.

c. To find the time at which only 20 mg of caffeine remains in the body, substitute \(y\) = 20 and solve for the corresponding value of \(t\).

\[\begin{array}{l} y=120(.87)^{t} \\ 20=120(.87)^{t} \end{array} \nonumber \]

Divide both sides by 120 to isolate the exponential expression.

\[\begin{array}{l} \frac{20}{120}=\frac{120}{120}\left(0.87^{t}\right) \\ 0.1667=0.87^{t} \end{array} \nonumber \]

Rewrite the expression in logarithmic form and use the change of base formula

\[\begin{array}{l} t=\log _{0.87}(0.1667) \\ t=\frac{\ln (0.1667)}{\ln (0.87)} \approx 12.9 \text { hours } \end{array} \nonumber \]

After 12.9 hours, 20 mg of caffeine remains in the body.

EXPRESSING EXPONENTIAL FUNCTIONS IN THE FORMS y = ab t and y = ae kt

Now that we’ve developed our equation solving skills, we revisit the question of expressing exponential functions equivalently in the forms \(y = ab^t\) and \(y = ae^{kt}\)

We’ve already determined that if given the form \(y = ae^{kt}\), it is straightforward to find \(b\).

Example \(\PageIndex{7}\)

For the following examples, assume \(t\) is measured in years.

  • Express \(y = 3500e^{0.25t}\) in form \(y = ab^t\) and find the annual percentage growth rate.
  • Express \(y = 28000e^{-0.32t}\) in form \(y = ab^t\) and find the annual percentage decay rate.

a. Express \(y = 3500e^{0.25t}\) in the form \(y = ab^t\)

\[\begin{array}{l} y=a e^{k t}=a b^{t} \\ a\left(e^{k}\right)^{t}=a b^{t} \end{array} \nonumber \]

Thus \(e^k=b\)

In this example \(b=e^{0.25} \approx 1.284\)

We rewrite the growth function as y = 3500(1.284 t )

To find \(r\), recall that \(b = 1+r\) \[\begin{aligned} &1.284=1+r\\ &0.284=\mathrm{r} \end{aligned} \nonumber \]

The continuous growth rate is \(k\) = 0.25 and the annual percentage growth rate is 28.4% per year.

b. Express \(y = 28000e^{-0.32t}\) in the form \(y = ab^t\)

In this example \(\mathrm{b}=e^{-0.32} \approx 0.7261\)

We rewrite the growth function as y = 28000(0.7261 t )

To find \(r\), recall that \(b = 1+r\) \[\begin{array}{l} 0.7261=1+r \\ 0.2739=r \end{array} \nonumber \]

The continuous decay rate is \(k\) = -0.32 and the annual percentage decay rate is 27.39% per year.

In the sentence, we omit the negative sign when stating the annual percentage decay rate because we have used the word “decay” to indicate that r is negative.

Example \(\PageIndex{8}\)

  • Express \(y = 4200 (1.078)^t\) in the form \(y =ae^{kt}\)
  • Express \(y = 150 (0.73)^t\) in the form \(y =ae^{kt}\)

a. Express \(y = 4200 (1.078)^t\) in the form \(y =ae^{kt}\)

\[\begin{array}{l} \mathrm{y}=\mathrm{a} e^{\mathrm{k} t}=\mathrm{ab}^{\mathrm{t}} \\ \mathrm{a}\left(e^{\mathrm{k}}\right)^{\mathrm{t}}=\mathrm{ab}^{\mathrm{t}} \\ e^{\mathrm{k}}=\mathrm{b} \\ e^{k}=1.078 \end{array} \nonumber \]

Therefore \(\mathrm{k}=\ln 1.078 \approx 0.0751\)

We rewrite the growth function as \(y = 3500e^{0.0751t}\)

b. Express \(y =150 (0.73)^t\) in the form \(y = ae^{kt}\)

\[\begin{array}{l} y=a e^{k t}=a b^{t} \\ a\left(e^{k}\right)^{t}=a b^{t} \\ e^{k}=b \\ e^{k}=0.73 \end{array} \nonumber \]

Therefore \(\mathrm{k}=\ln 0.73 \approx-0.3147\)

We rewrite the growth function as \(y = 150e^{-0.3147t}\)

AN APPLICATION OF A LOGARITHMIC FUNCTON

Suppose we invest $10,000 today and want to know how long it will take to accumulate to a specified amount, such as $15,000. The time \(t\) needed to reach a future value \(y\) is a logarithmic function of the future value: \(t = g(y)\)

Example \(\PageIndex{9}\)

Suppose that Vinh invests $10000 in an investment earning 5% per year. He wants to know how long it would take his investment to accumulate to $12000, and how long it would take to accumulate to $15000.

We start by writing the exponential growth function that models the value of this investment as a function of the time since the $10000 is initially invested

\[y=10000(1.05)^{t} \nonumber \]

We divide both sides by 10000 to isolate the exponential expression on one side.

\[\frac{y}{10000}=1.05^{t} \nonumber \]

Next we rewrite this in logarithmic form to express time as a function of the accumulated future value. We’ll use function notation and call this function \(g(y)\).

\[\mathrm{t}=\mathrm{g}(\mathrm{y})=\log _{1.05}\left(\frac{\mathrm{y}}{10000}\right) \nonumber \]

Use the change of base formula to express \(t\) as a function of \(y\) using natural logarithm:

\[\mathrm{t}=\mathrm{g}(\mathrm{y})=\frac{\ln \left(\frac{\mathrm{y}}{10000}\right)}{\ln (1.05)} \nonumber \]

We can now use this function to answer Vinh’s questions.

To find the number of years until the value of this investment is $12,000, we substitute \(y\) = $12,000 into function \(g\) and evaluate \(t\):

\[\mathrm{t}=\mathrm{g}(12000)=\frac{\ln \left(\frac{12000}{10000}\right)}{\ln (1.05)}=\frac{\ln (1.2)}{\ln (1.05)}=3.74 \text { years } \nonumber \]

To find the number of years until the value of this investment is $15,000, we substitute \(y\) = $15,000 into function \(g\) and evaluate \(t\):

\[\mathrm{t}=\mathrm{g}(15000)=\frac{\ln \left(\frac{15000}{10000}\right)}{\ln (1.05)}=\frac{\ln (1.5)}{\ln (1.05)}=8.31 \text { years } \nonumber \]

Before ending this section, we investigate the graph of the function \(\mathrm{t}=\mathrm{g}(\mathrm{y})=\frac{\ln \left(\frac{\mathrm{y}}{10000}\right)}{\ln (1.05)}\). We see that the function has the general shape of logarithmic functions that we examined in section 5.5. From the points plotted on the graph, we see that function \(g\) is an increasing function but it increases very slowly.

5.6.png

If we consider just the function \(\mathrm{t}=\mathrm{g}(\mathrm{y})=\frac{\ln \left(\frac{\mathrm{y}}{10000}\right)}{\ln (1.05)}\), then the domain of function would be \(y > 0\), all positive real numbers, and the range for \(t\) would be all real numbers.

In the context of this investment problem, the initial investment at time \(t\) = 0 is \(y\) =$10,000. Negative values for time do not make sense. Values of the investment that are lower than the initial amount of $10,000 also do not make sense for an investment that is increasing in value.

Therefore the function and graph as it pertains to this problem concerning investments has domain \(y ≥ 10,000\) and range \(t ≥ 0\).

The graph below is restricted to the domain and range that make practical sense for the investment in this problem.

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COMMENTS

  1. Algebra

    For problems 1 - 3 write the expression in logarithmic form. 75 =16807 7 5 = 16807 Solution 163 4 = 8 16 3 4 = 8 Solution (1 3)−2 = 9 ( 1 3) − 2 = 9 Solution For problems 4 - 6 write the expression in exponential form. log232 = 5 log 2 32 = 5 Solution log1 5 1 625 = 4 log 1 5 1 625 = 4 Solution log9 1 81 = −2 log 9 1 81 = − 2 Solution

  2. Solving Logarithmic Equations

    Example 1: Solve the logarithmic equation. Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it. Here is the rule, just in case you forgot. Given Apply Product Rule from Log Rules. Distribute: [latex]\left ( {x + 2} \right)\left ( 3 \right) = 3x + 6 [/latex]

  3. Evaluate logarithms (practice)

    Do 4 problems. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

  4. Logarithms

    Familiar Attempted Not started Quiz Unit test About this unit Logarithms are the inverses of exponents. They allow us to solve challenging exponential equations, and they are a good excuse to dive deeper into the relationship between a function and its inverse. Introduction to logarithms Learn Intro to logarithms Intro to Logarithms

  5. Algebra

    Example 1 Solve each of the following equations. 2log9(√x)−log9(6x−1) = 0 2 log 9 ( x) − log 9 ( 6 x − 1) = 0 logx+log(x −1) = log(3x +12) log x + log ( x − 1) = log ( 3 x + 12) ln10−ln(7−x) = lnx ln 10 − ln ( 7 − x) = ln x Show All Solutions Hide All Solutions a 2log9(√x)−log9(6x −1) = 0 2 log 9 ( x) − log 9 ( 6 x − 1) = 0 Show Solution

  6. Intro to Logarithms (article)

    The answer would be 4 . This is expressed by the logarithmic equation log 2 ( 16) = 4 , read as "log base two of sixteen is four". 2 4 = 16 log 2 ( 16) = 4. Both equations describe the same relationship between the numbers 2 , 4 , and 16 , where 2 is the base and 4 is the exponent. The difference is that while the exponential form isolates the ...

  7. Solving Logarithmic Equations

    Example: The logarithm of the number 1 to any non-zero base is always zero. b 0 =1 log b 1 = 0. How to Solve Logarithmic Equations? An equation containing variables in the exponents is knowns as an exponential equation.

  8. Log problems: pH, decibels, and the Richter Scale

    The classic real-world contexts for logarithm word problems are the measurement of acidity or alkalinity (that is, the measurement of pH), the measurement of sound (in decibels, or dB), and the measurement of earthquake intensity (on the Richter scale), among other uses ( link ).

  9. Logarithms

    A logarithm is the inverse of the exponential function. Specifically, a logarithm is the power to which a number (the base) must be raised to produce a given number. For example, \log_2 64 = 6, log2 64 = 6, because 2^6 = 64. 26 = 64. In general, we have the following definition: z z is the base- x x logarithm of y y if and only if x^z = y xz = y.

  10. Logarithm (Logs)

    An exponential equation is converted into a logarithmic equation and vice versa using b x = a ⇔ log b a = x. A common log is a logarithm with base 10, i.e., log 10 = log. A natural log is a logarithm with base e, i.e., log e = ln. Logarithms are used to do the most difficult calculations of multiplication and division.

  11. 4.5e: Exercises

    Answers to odd exercises: ★ In the following exercises, use the Properties of Logarithms to expand the logarithm. Simplify if possible. 111. log3 3√x2 27y4. 112. log4 √x 16y4. 113. log3√3x + 2y2 5z2. 114. log2√2x + y2 8z2.

  12. Logarithmic Equations

    EXAMPLE 1 What is the result of \log_ {5} (x+1)+\log_ {5} (3)=\log_ {5} (15) log5(x + 1) + log5(3) = log5(15)? Solution EXAMPLE 2 Solve the equation \log_ {4} (2x+2)+\log_ {4} (2)=\log_ {4} (x+1)+\log_ {4} (3) log4(2x + 2) + log4(2) = log4(x+ 1) + log4(3) Solution EXAMPLE 3

  13. Solving Logarithmic Functions

    12 2 = 144. log 12 144 = 2. log base 12 of 144. Let's use these properties to solve a couple of problems involving logarithmic functions. Example 1. Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function. Solution. Given 7 2 = 64. Here, the base = 7, exponent = 2 and the argument = 49.

  14. Logarithmic Functions (video lessons, examples and solutions)

    When given a problem on solving a logarithmic equation with multiple logs, students should understand how to condense logarithms. By condensing the logarithms, we can create an equation with only one log, and can use methods of exponentiation for solving a logarithmic equation with multiple logs. ... More examples on solving logarithmic ...

  15. Algebra

    Example 1 Evaluate each of the following logarithms. log416 log 4 16 log216 log 2 16 log6216 log 6 216 log5 1 125 log 5 1 125 log1 381 log 1 3 81 log3 2 27 8 log 3 2 27 8 Show All Solutions Hide All Solutions Show Discussion a log416 log 4 16 Show Solution b log216 log 2 16 Show Solution c log6216 log 6 216 Show Solution

  16. Introduction to Logarithms

    Sometimes a logarithm is written without a base, like this: log (100) This usually means that the base is really 10. It is called a "common logarithm". Engineers love to use it. On a calculator it is the "log" button. It is how many times we need to use 10 in a multiplication, to get our desired number. Example: log (1000) = log10(1000) = 3.

  17. Evaluating logarithms (advanced) (video)

    To understand the reason why log(1023) equals approximately 3.0099 we have to look at how logarithms work. Saying log(1023) = 3.009 means 10 to the power of 3.009 equals 1023. The ten is known as the base of the logarithm, and when there is no base, the default is 10. 10^3 equals 1000, so it makes sense that to get 1023 you have to put 10 to ...

  18. 4.6e: Exercises

    Exercise 4.6e. 5. ★ For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation. 121. log( 1 100) = − 2. 122. log324(18) = 1 2. ★ For the following exercises, use the definition of a logarithm to solve the equation. 123. 5log7n = 10.

  19. 6.6: Exponential and Logarithmic Equations

    Solve the resulting equation, S = T. S = T. , for the unknown. Example 6.6.1: Solving an Exponential Equation with a Common Base. Solve 2x − 1 = 22x − 4. Solution. 2x − 1 = 22x − 4 The common base is 2 x − 1 = 2x − 4 By the one-to-one property the exponents must be equal x = 3 Solve for x. Exercise 6.6.1.

  20. Algebra

    Solve each of the following equations. log4(x2−2x) = log4(5x −12) log 4 ( x 2 − 2 x) = log 4 ( 5 x − 12) Solution log(6x) −log(4−x) =log(3) log ( 6 x) − log ( 4 − x) = log ( 3) Solution ln(x)+ln(x+3) =ln(20−5x) ln ( x) + ln ( x + 3) = ln ( 20 − 5 x) Solution log3(25−x2) = 2 log 3 ( 25 − x 2) = 2 Solution

  21. Common and Natural Logarithm (video lessons, examples and solutions)

    The natural logarithm has base e, a famous irrational number, and is represented on the calculator by ln(x). The natural and common logarithm can be found throughout Algebra and Calculus. Defines common log, log x, and natural log, ln x, and works through examples and problems using a calculator. Show Video Lesson

  22. Logarithm Questions

    The questions of logarithm could be solved based on the properties, given below: Product rule: log b MN = log b M + log b N Quotient rule: log b M/N = log b M - log b N Power rule: log b M p = P log b M Zero Exponent Rule: log a 1 = 0 Change of Base Rule: log b (x) = ln x / ln b or log b (x) = log 10 x / log 10 b Also, read: Logarithms

  23. Solving Equations

    For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better. Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying ...

  24. 5.6: Application Problems with Exponential and Logarithmic Functions

    We divide both sides by 10000 to isolate the exponential expression on one side. y 10000 = 1.05t y 10000 = 1.05 t. Next we rewrite this in logarithmic form to express time as a function of the accumulated future value. We'll use function notation and call this function g(y) g ( y).

  25. How to Use Problem-Solving Skills in Graphic Design Interviews

    Learn how to showcase your problem-solving skills in graphic design interviews with examples of analytical, creative, critical, collaborative, strategic, and reflective thinking.

  26. Find the AI Approach That Fits the Problem You're Trying to Solve

    AI moves quickly, but organizations change much more slowly. What works in a lab may be wrong for your company right now. If you know the right questions to ask, you can make better decisions ...