Exponential and Logarithmic Functions: Learn It 3

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Exponential and Logarithmic Functions: Learn It 3

Logarithmic Functions

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.

Inverse Functions

For any one-to-one function $f\left(x\right)=y$ , a function ${f}^{-1}\left(x\right)$ is an inverse function of $f$ if ${f}^{-1}\left(y\right)=x$ .

The notation ${f}^{-1}$ is read “ $f$ inverse.” Like any other function, we can use any variable name as the input for ${f}^{-1}$ , so we will often write ${f}^{-1}\left(x\right)$ , which we read as $"f$ inverse of $x$ “.

Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.

The exponential function $f(x)=b^x$ is one-to-one, with domain $(−\infty ,\infty)$ and range $(0,\infty )$ . Therefore, it has an inverse function, called the logarithmic function with base $b$ .

For any $b>0, \, b \ne 1$ , the logarithmic function with base $b$ , denoted $\log_b$ , has domain $(0,\infty )$ and range $(−\infty ,\infty )$ , and satisfies

$\log_b(x)=y$ if and only if $b^y=x$ .

logarithmic functions

A logarithmic function is the inverse of an exponential function and is written as $log_{b}(x)$ . For a given base $b$ , it tells us the power to which $b$ must be raised to get $x$ .

\begin{matrix} {log}_{2} (8) = 3 & since 2^{3} = 8, {log}_{10} (\frac{1}{100}) = - 2 & since 10^{- 2} = \frac{1}{10^{2}} = \frac{1}{100}, {log}_{b} (1) = 0 & since b^{0} = 1 for any base b > 0. \end{matrix}

$\begin{array}{cccc} \log_2 (8)=3\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}2^3=8,\hfill \\ \log_{10} (\frac{1}{100})=-2\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}10^{-2}=\frac{1}{10^2}=\frac{1}{100},\hfill \\ \log_b (1)=0\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}b^0=1 \, \text{for any base} \, b>0.\hfill \end{array}$

The most commonly used logarithmic function is the function $\log_e (x)$ . Since this function uses natural $e$ as its base, it is called the natural logarithm. Here we use the notation $\ln(x)$ or $\ln x$ to mean $\log_e (x)$ .

\begin{matrix} ln (e) = {log}_{e} (e) = 1 ln (e^{3}) = {log}_{e} (e^{3}) = 3 ln (1) = {log}_{e} (1) = 0 \end{matrix}

$\begin{array}{l}\ln (e)=\log_e (e)=1 \\ \ln(e^3)=\log_e (e^3)=3 \\ \ln(1)=\log_e (1)=0\end{array}$

Euler’s number, denoted as $e$ , is a fundamental mathematical constant approximately equal to $2.71828$ . It is the base of the natural logarithm and the natural exponential function, known for its unique properties in calculus, especially in relation to growth processes and compound interest calculations.

Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.

Properties of Logarithms

If $a,b,c>0, \, b\ne 1$ , and $r$ is any real number, then

$\begin{array}{cccc}1.\phantom{\rule{2em}{0ex}}\log_b (ac)=\log_b (a)+\log_b (c)\hfill & & & \text{(Product property)}\hfill \\ 2.\phantom{\rule{2em}{0ex}}\log_b(\frac{a}{c})=\log_b (a) -\log_b (c)\hfill & & & \text{(Quotient property)}\hfill \\ 3.\phantom{\rule{2em}{0ex}}\log_b (a^r)=r \log_b (a)\hfill & & & \text{(Power property)}\hfill \end{array}$

Solve each of the following equations for $x$ .

$\ln \left(\frac{1}{x}\right)=4$
$\log_{10} \sqrt{x}+ \log_{10} x=2$
$\ln(2x)-3 \ln(x^2)=0$

By the definition of the natural logarithm function,
$\ln\big(\frac{1}{x}\big)=4 \, \text{ if and only if } \, e^4=\frac{1}{x}$

Therefore, the solution is $x=\frac{1}{e^4}$ .
Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
$\log_{10} \sqrt{x}+ \log_{10} x = \log_{10} x \sqrt{x} = \log_{10}x^{3/2} = \frac{3}{2} \log_{10} x$

Therefore, the equation can be rewritten as

$\frac{3}{2} \log_{10} x = 2 \, \text{ or } \, \log_{10} x = \frac{4}{3}$

The solution is $x=10^{4/3}=10\sqrt[3]{10}$ .
Using the power property of logarithmic functions, we can rewrite the equation as $\ln(2x) - \ln(x^6) = 0$ .
Using the quotient property, this becomes

$\ln\big(\frac{2}{x^5}\big)=0$

Therefore, $\frac{2}{x^5}=1$ , which implies $x=\sqrt[5]{2}$ .

Solve each of the following equations for $x$ .

$5^x=2$
$e^x+6e^{−x}=5$

Applying the natural logarithm function to both sides of the equation, we have
$\ln 5^x=\ln 2$

Using the power property of logarithms,

$x \ln 5=\ln 2$

Therefore, $x=\frac{\ln 2 }{\ln 5}$ .
Multiplying both sides of the equation by $e^x$ , we arrive at the equation
$e^{2x}+6=5e^x$

Rewriting this equation as

$e^{2x}-5e^x+6=0$ ,

we can then rewrite it as a quadratic equation in $e^x$ :

$(e^x)^2-5(e^x)+6=0$

Now we can solve the quadratic equation. Factoring this equation, we obtain

$(e^x-3)(e^x-2)=0$

Therefore, the solutions satisfy $e^x=3$ and $e^x=2$ . Taking the natural logarithm of both sides gives us the solutions

$x=\ln 3, \, \ln 2$

Watch the following video to see the worked solution to this example.

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You can view the transcript for this video using this link (opens in new window).

In calculations involving logarithms, you might have noticed that calculators typically provide only the common logarithm ( $\log_{10}$ ) and natural logarithm (base $e$ ). However, exponential functions and logarithm functions can be expressed in terms of any desired base $b$ .

To work with different bases, we can use the change-of-base formula to convert to a base your calculator can handle.

change-of-base formulas

The change-of-base formula allows you to evaluate logarithms with any base using only the common or natural logarithm functions typically available on a calculator.

For any real number $x$ and bases $a>0, \, b>0$ , and $a\ne 1, \, b\ne 1$ , the exponential expression $a^x$ can be rewritten using a logarithm with base $b$ as $a^x=b^{x \log_b a}$ .

The change of base formula is:

{log}_{a} x = \frac{{log}_{b} x}{{log}_{b} a}

$\log_a x=\frac{\log_b x}{\log_b a}$ for any real number

x > 0

$x>0$ .

If $b=e$ , this exponential expression reduces to $a^x=e^{x \log_e a}=e^{x \ln a}$ . The change of base formula 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 $b>0, \, b\ne 1, \, \log_b (a^x)=x \log_b a$ . Therefore,

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

$b^{\log_b(a^x)}=b^{x \log_b a}$

In addition, we know that $b^x$ and $\log_b (x)$ are inverse functions. Therefore,

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

$b^{\log_b (a^x)}=a^x$

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

$(\log_b a)·(\log_a x)=\log_b x$

Let $u=\log_b a, \, v=\log_a x$ , and $w=\log_b x$ . We will show that $u·v=w$ . 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}

$b^{uv}=(b^u)^v=a^v=x=b^w$

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

$_\blacksquare$

How to: Use the Change-of-Base Formulas

Take the logarithm you need to evaluate, $(\log_b a)$
Using the change-of-base formula, rewrite it as:
$\log_a x=\frac{\log_b x}{\log_b a}$
Use your calculator to find the common log (base $10$ ) of $a$ , $\log{a}$ , and the common log of $x$ , $\log{x}$ .
Divide these two values to compute $\log_a x$

Note: You can also use the natural logarithm ( $ln$ ) in place of the common logarithm ( $log$ ) if preferred.

Use a calculator to evaluate $\log_3 7$ using the change-of-base formula.

Using the change-of-base formula $\log_a x=\frac{\log_b x}{\log_b a}$ :

$\log_3 7=\frac{\log 7}{\log 3} \approx 1.77124$ .