# 4.3 Logarithmic functions  (Page 2/9)

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We can express the relationship between logarithmic form and its corresponding exponential form as follows:

${\mathrm{log}}_{b}\left(x\right)=y⇔{b}^{y}=x,\text{}b>0,b\ne 1$

Note that the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is always positive.

Because logarithm is a function, it is most correctly written as $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(x\right),$ using parentheses to denote function evaluation, just as we would with $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as $\text{\hspace{0.17em}}{\mathrm{log}}_{b}x.\text{\hspace{0.17em}}$ Note that many calculators require parentheses around the $\text{\hspace{0.17em}}x.$

We can illustrate the notation of logarithms as follows:

Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y={b}^{x}\text{\hspace{0.17em}}$ are inverse functions.

## Definition of the logarithmic function

A logarithm    base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ of a positive number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ satisfies the following definition.

For $\text{\hspace{0.17em}}x>0,b>0,b\ne 1,$

where,

• we read $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ as, “the logarithm with base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}x$ ” or the “log base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}x."$
• the logarithm $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the exponent to which $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ must be raised to get $\text{\hspace{0.17em}}x.$

Also, since the logarithmic and exponential functions switch the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,

• the domain of the logarithm function with base
• the range of the logarithm function with base

Can we take the logarithm of a negative number?

No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.

Given an equation in logarithmic form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(x\right)=y,$ convert it to exponential form.

1. Examine the equation $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}x\text{\hspace{0.17em}}$ and identify $\text{\hspace{0.17em}}b,y,\text{and}x.$
2. Rewrite $\text{\hspace{0.17em}}{\mathrm{log}}_{b}x=y\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{b}^{y}=x.$

## Converting from logarithmic form to exponential form

Write the following logarithmic equations in exponential form.

1. ${\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}$
2. ${\mathrm{log}}_{3}\left(9\right)=2$

First, identify the values of $\text{\hspace{0.17em}}b,y,\text{and}x.\text{\hspace{0.17em}}$ Then, write the equation in the form $\text{\hspace{0.17em}}{b}^{y}=x.$

1. ${\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}$

Here, Therefore, the equation $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{6}^{\frac{1}{2}}=\sqrt{6}.$

2. ${\mathrm{log}}_{3}\left(9\right)=2$

Here, Therefore, the equation $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(9\right)=2\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{3}^{2}=9.\text{\hspace{0.17em}}$

Write the following logarithmic equations in exponential form.

1. ${\mathrm{log}}_{10}\left(1,000,000\right)=6$
2. ${\mathrm{log}}_{5}\left(25\right)=2$
1. ${\mathrm{log}}_{10}\left(1,000,000\right)=6\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{10}^{6}=1,000,000$
2. ${\mathrm{log}}_{5}\left(25\right)=2\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{5}^{2}=25$

## Converting from exponential to logarithmic form

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base $\text{\hspace{0.17em}}b,$ exponent $\text{\hspace{0.17em}}x,$ and output $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Then we write $\text{\hspace{0.17em}}x={\mathrm{log}}_{b}\left(y\right).$

## Converting from exponential form to logarithmic form

Write the following exponential equations in logarithmic form.

1. ${2}^{3}=8$
2. ${5}^{2}=25$
3. ${10}^{-4}=\frac{1}{10,000}$

First, identify the values of $\text{\hspace{0.17em}}b,y,\text{and}x.\text{\hspace{0.17em}}$ Then, write the equation in the form $\text{\hspace{0.17em}}x={\mathrm{log}}_{b}\left(y\right).$

1. ${2}^{3}=8$

Here, $\text{\hspace{0.17em}}b=2,$ $\text{\hspace{0.17em}}x=3,$ and $\text{\hspace{0.17em}}y=8.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{2}^{3}=8\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(8\right)=3.$

2. ${5}^{2}=25$

Here, $\text{\hspace{0.17em}}b=5,$ $\text{\hspace{0.17em}}x=2,$ and $\text{\hspace{0.17em}}y=25.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{5}^{2}=25\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(25\right)=2.$

3. ${10}^{-4}=\frac{1}{10,000}$

Here, $\text{\hspace{0.17em}}b=10,$ $\text{\hspace{0.17em}}x=-4,$ and $\text{\hspace{0.17em}}y=\frac{1}{10,000}.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{10}^{-4}=\frac{1}{10,000}\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\text{log}}_{10}\left(\frac{1}{10,000}\right)=-4.$

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