1.5 Exponential and logarithmic functions (Page 5/17)

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Solve $ln (x^{3}) - 4 ln (x) = 1 .$

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

When evaluating a logarithmic function with a calculator, you may have noticed that the only options are ${log}_{10}$ 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 $a \neq 1, b \neq 1 .$

$a^{x} = b^{x {log}_{b} a}$ for any real number $x .$
If $b = e,$ this equation reduces to $a^{x} = e^{x {log}_{e} a} = e^{x ln a} .$
${log}_{a} x = \frac{{log}_{b} x}{{log}_{b} a}$ for any real number $x > 0 .$
If $b = e,$ this equation reduces to ${log}_{a} x = \frac{ln x}{ln a} .$

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 $b > 0, b \neq 1, {log}_{b} (a^{x}) = x {log}_{b} a .$ Therefore,

b^{{log}_{b} (a^{x})} = b^{x {log}_{b} a} .

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

b^{{log}_{b} (a^{x})} = a^{x} .

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

To prove the second property, we show that

({log}_{b} a) \cdot ({log}_{a} x) = {log}_{b} x .

Let $u = {log}_{b} a, v = {log}_{a} x,$ and $w = {log}_{b} x .$ We will show that $u \cdot v = w .$ 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^{u v} = {(b^{u})}^{v} = a^{v} = x = b^{w} .

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

□

Changing bases

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

Use the second equation with $a = 3$ and $e = 3 :$

${log}_{3} 7 = \frac{ln 7}{ln 3} \approx 1.77124 .$

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

$1.29248$

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Chapter opener: the richter scale for earthquakes

A photograph of an earthquake fault. — (credit: modification of work by Robb Hannawacker, NPS)

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} = {log}_{10} (\frac{A_{1}}{A_{2}}) .

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

8 - 7 = {log}_{10} (\frac{A_{1}}{A_{2}}) .

Therefore,

{log}_{10} (\frac{A_{1}}{A_{2}}) = 1,

which implies $A_{1} / A_{2} = 10$ 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

{log}_{10} (\frac{A_{1}}{A_{2}}) = 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:

$9 - 7.3 = {log}_{10} (\frac{A_{1}}{A_{2}}) .$

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

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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