# 1.5 Exponential and logarithmic functions  (Page 5/17)

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Solve $\text{ln}\left({x}^{3}\right)-4\phantom{\rule{0.1em}{0ex}}\text{ln}\left(x\right)=1.$

$x=\frac{1}{e}$

When evaluating a logarithmic function with a calculator, you may have noticed that the only options are ${\text{log}}_{10}$ or log, called the common logarithm , or ln , which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base $b.$ If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.

## Rule: change-of-base formulas

Let $a>0,b>0,$ and $a\ne 1,b\ne 1.$

1. ${a}^{x}={b}^{x{\text{log}}_{b}a}$ for any real number $x.$
If $b=e,$ this equation reduces to ${a}^{x}={e}^{x{\text{log}}_{e}a}={e}^{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}a}.$
2. ${\text{log}}_{a}x=\frac{{\text{log}}_{b}x}{{\text{log}}_{b}a}$ for any real number $x>0.$
If $b=e,$ this equation reduces to ${\text{log}}_{a}x=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}a}.$

## Proof

For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base $b>0,b\ne 1,{\text{log}}_{b}\left({a}^{x}\right)=x{\text{log}}_{b}a.$ Therefore,

${b}^{{\text{log}}_{b}\left({a}^{x}\right)}={b}^{x{\text{log}}_{b}a}.$

In addition, we know that ${b}^{x}$ and ${\text{log}}_{b}\left(x\right)$ are inverse functions. Therefore,

${b}^{{\text{log}}_{b}\left({a}^{x}\right)}={a}^{x}.$

Combining these last two equalities, we conclude that ${a}^{x}={b}^{x{\text{log}}_{b}a}.$

To prove the second property, we show that

$\left({\text{log}}_{b}a\right)·\left({\text{log}}_{a}x\right)={\text{log}}_{b}x.$

Let $u={\text{log}}_{b}a,v={\text{log}}_{a}x,$ and $w={\text{log}}_{b}x.$ We will show that $u·v=w.$ By the definition of logarithmic functions, we know that ${b}^{u}=a,{a}^{v}=x,$ and ${b}^{w}=x.$ From the previous equations, we see that

${b}^{uv}={\left({b}^{u}\right)}^{v}={a}^{v}=x={b}^{w}.$

Therefore, ${b}^{uv}={b}^{w}.$ Since exponential functions are one-to-one, we can conclude that $u·v=w.$

## Changing bases

Use a calculating utility to evaluate ${\text{log}}_{3}7$ with the change-of-base formula presented earlier.

Use the second equation with $a=3$ and $e=3\text{:}$

${\text{log}}_{3}7=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}7}{\text{ln}\phantom{\rule{0.1em}{0ex}}3}\approx 1.77124.$

Use the change-of-base formula and a calculating utility to evaluate ${\text{log}}_{4}6.$

$1.29248$

## Chapter opener: the richter scale for earthquakes

In 1935, Charles Richter developed a scale (now known as the Richter scale ) to measure the magnitude of an earthquake . The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude ${R}_{1}$ on the Richter scale and a second earthquake with magnitude ${R}_{2}$ on the Richter scale. Suppose ${R}_{1}>{R}_{2},$ which means the earthquake of magnitude ${R}_{1}$ is stronger, but how much stronger is it than the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If ${A}_{1}$ is the amplitude measured for the first earthquake and ${A}_{2}$ is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:

${R}_{1}-{R}_{2}={\text{log}}_{10}\left(\frac{{A}_{1}}{{A}_{2}}\right).$

Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,

$8-7={\text{log}}_{10}\left(\frac{{A}_{1}}{{A}_{2}}\right).$

Therefore,

${\text{log}}_{10}\left(\frac{{A}_{1}}{{A}_{2}}\right)=1,$

which implies ${A}_{1}\text{/}{A}_{2}=10$ or ${A}_{1}=10{A}_{2}.$ Since ${A}_{1}$ is 10 times the size of ${A}_{2},$ we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation

${\text{log}}_{10}\left(\frac{{A}_{1}}{{A}_{2}}\right)=8-6=2.$

Therefore, ${A}_{1}=100{A}_{2}.$ That is, the first earthquake is 100 times more intense than the second earthquake.

How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010?

To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:

$9-7.3={\text{log}}_{10}\left(\frac{{A}_{1}}{{A}_{2}}\right).$

Therefore, ${A}_{1}\text{/}{A}_{2}={10}^{1.7},$ and we conclude that the earthquake in Japan was approximately $50$ times more intense than the earthquake in Haiti.

#### Questions & Answers

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