The Inverse of the Logarithmic Function

Graph the inverse of the following logarithmic function. ||y = -6\log_5 (x+4)+3||

1. Graph the logarithmic function in order to graph the inverse

2. Draw the line |y = x|

3. Reflect the starting logarithmic function in terms of the line |y = x|

Therefore, we obtain the inverse of the starting logarithmic function.

1. Swap the variables |x| and |y| in the initial rule.

2. Isolate the expression containing the logarithm.

3. Switch to exponential form to isolate |y.|

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

1. Swap the |x| and |y| variables in the initial rule

   ||x = -4\log_7 (3(y-6))+8||

2. Isolate the expression containing the logarithm

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

3. Switch to the exponential form to isolate |y|

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

When carefully observing the starting function and its inverse, note the following:

• The parameter |h| becomes the parameter |k| of the inverse;

• The parameter |k| becomes the parameter |h| of the inverse;

• The inverse’s base |c| is the same as the starting function’s base;

• The inverse’s parameter |a| is the reciprocal of the starting function’s parameter |b;|

• The inverse’ s parameter |b| is the reciprocal of the starting function’s parameter |a.|

||y = \color{red}{a}\log_\color{magenta}{c} \big(\color{purple}{b}(x-\color{blue}{h})\big)+\color{green}{k}\ \ \Leftrightarrow \ \ y^{-1}=\color{purple}{\dfrac{1}{b}}(\color{magenta}{c})^{\color{red}{\dfrac{1}{a}}(x-\color{green}{k})}+\color{blue}{h}||