The Inverse of the Logarithmic Function

1. Graph the function using points.
2. Draw the line of reflection |y=x.|
3. Reflect a few points of the function across the line |y=x.|
4. Graph the inverse.

Graph the inverse of the following logarithmic function: ||y = -3\log_2\left(-\dfrac{1}{4}x\right)||

1. Graph the function using points.

Since there are no given points or table of values, we must use the rule and substitute |x| with any value.
If |x = -8|, we get: ||\begin{align}f(-8) &= -3\log_2\left(-\dfrac{1}{4}\times-8\right)\\
&= -3\log_2\left(2\right)\\
&= -3\times1\\
&= -3\end{align}||
The graph of the logarithmic function therefore passes through point |(-8, -3).| We repeat this for other values of |x| to find other points.

2. Draw the line of reflection |\boldsymbol{y = x}|.

1. Reflect the function across the line |\boldsymbol {y=x}|.

We interchange the |x| and |y| coordinates of the points, then plot them on the graph.

4. Graph the function

Therefore, we get the inverse of the initial logarithmic function. We can use the points of the inverse to find the rule of this exponential function.

A logarithmic function and its inverse always have the same variation.

1. Interchange the variables |x| and |y| in the initial rule.
2. Isolate the expression containing the logarithm.
3. Isolate |y| by converting to exponential form.

Determine algebraically the rule of the inverse of the following logarithmic function.
||f(x)=-4\log_7\left(3(x-6)\right)+8||

1. Interchange the variables |\boldsymbol{x}| and |\boldsymbol{y}| in the initial rule.

||\begin{gather}\boldsymbol{y}=-4\log_7\left(3(\boldsymbol{x}-6)\right)+8
\\\Downarrow\\
\boldsymbol{x}=-4\log_7\left(3(\boldsymbol{y}-6)\right)+8\end{gather}||

2. Isolate the expression containing the logarithm.

||\begin{align}x &= -4\log_7\left(3(y-6)\right)+8\\
x-8 &= -4\log_7\left(3(y-6)\right)\\
-\dfrac{1}{4}(x-8) &= \log_7\left(3(y-6)\right)\end{align}||

3. Convert to exponential form to isolate |\boldsymbol{y}.|

||\begin{align}7^{\frac{-1}{4}(x\,-\,8)} &= 3(y-6)\\ \dfrac{7^{\frac{-1}{4}(x\,-\,8)}}{3} &= y-6\\  \dfrac{7^{\frac{-1}{4}(x\,-\,8)}}{3}+6 &= y\\
\dfrac{1}{3}(7)^{\frac{-1}{4}(x\,-\,8)}+6 &= y\end{align}||
Therefore, the rule of the inverse is |f^{-1}(x)=\dfrac{1}{3}(7)^{\frac{-1}{4}(x\,-\,8)}+6.|
 

If we carefully observe the initial function |\left(f(x)\right)| and its inverse |\left(f^{-1}(x)\right),| this is what we notice:

• The base |\boldsymbol{\color{#FF55C3}{c}}| of the inverse is the same as that of the initial function.
• The inverse of parameter |\boldsymbol{\color{#3A9A38}{a}}| of the initial function corresponds to parameter |\boldsymbol{\color{#EC0000}{b}}| of the inverse.
• The inverse of parameter |\boldsymbol{\color{#EC0000}{b}}| of the initial function corresponds to parameter |\boldsymbol{\color{#3A9A38}{a}}| of the inverse.
• Parameter |\boldsymbol{\color{#3B87CD}{h}}| corresponds to parameter |\boldsymbol{\color{#FA7921}{k}}| of the inverse.
• Parameter |\boldsymbol{\color{#FA7921}{k}}| corresponds to parameter |\boldsymbol{\color{#3B87CD}{h}}| of the inverse.

||f(x)=\boldsymbol{\color{#3A9A38}{a}}\log_\boldsymbol{\color{#FF55C3}{c}}\left(\boldsymbol{\color{#EC0000}{b}}(x-\boldsymbol{\color{#3B87CD}{h}})\right)+\boldsymbol{\color{#FA7921}{k}}\\[5pt]
\Updownarrow\\[5pt]
f^{-1}(x)=\boldsymbol{\color{#3A9A38}{\dfrac{1}{b}}}(\boldsymbol{\color{#FF55C3}{c}})^{\boldsymbol{\color{#EC0000}{\frac{1}{a}}}(x\,-\,\boldsymbol{\color{#3B87CD}{k}})}+\boldsymbol{\color{#FA7921}{h}}||

Find the rule of the inverse of the following logarithmic function:
||f(x)=0.25\log_{10}\left(-\dfrac{2}{7}(x+9)\right)-6||

We can directly find the inverse as follows.
||\begin{gather}f(x)=\boldsymbol{\color{#3A9A38}{0.25}}\log_\boldsymbol{\color{#FF55C3}{10}} \left(\boldsymbol{\color{#EC0000}{-\dfrac{2}{7}}}(x-\boldsymbol{\color{#3B87CD}{-9}})\right)+\boldsymbol{\color{#FA7921}{-6}}\\[3pt]
\Updownarrow \\[3pt]
\begin{aligned}f^{-1}(x)&=\boldsymbol{\color{#3A9A38}{\dfrac{1}{-\frac{2}{7}}}}(\boldsymbol{\color{#FF55C3}{10}})^{\boldsymbol{\color{#EC0000}{\frac{1}{0.25}}}(x\,-\,\boldsymbol{\color{#3B87CD}{-6})}}+\boldsymbol{\color{#FA7921}{-9}}
\\[5pt] &= -\dfrac{7}{2}(10)^{4(x\,+\,6)}-9\end{aligned}\end{gather}||