Inverses of Exponential & Logarithmic Functions (College Board AP® Precalculus): Revision Note

Inverses of exponential functions

How do I find the inverse of a transformed exponential function?

• A transformed exponential function has the general form

  • $f\left(x\right)=a{b}^{(x+h)}+k$

• This is a combination of

  • additive transformations

    • horizontal shift $h$

    • vertical shift $k$

  • and a vertical dilation

    • factor $a$

  • applied to the base exponential function $b^x$

• To find the inverse, think of the function as a sequence of operations applied to the input $x$

  • and then reverse each operation in the opposite order

  • Make sure you are familiar with the process for finding an inverse function

• In general this will proceed as follows (using $y=a{b}^{(x+h)}+k$)

  • Start with the equation

    • $y=a{b}^{(x+h)}+k$

  • Subtract $k$

    • $y-k=a{b}^{(x+h)}$

  • Divide by $a$

    • $\frac{y-k}{a}=b^{(x+h)}$

  • Take ${\log }_b$ of both sides (this 'cancels' the exponential on the right)

    • ${\log }_b\left(\frac{y-k}{a}\right)=x+h$

  • Subtract $h$

    • $x={\log }_b\left(\frac{y-k}{a}\right)-h$

  • Swap $x$ and $y$

    • $f^{-1}\left(x\right)={\log }_b\left(\frac{x-k}{a}\right)-h$

• The inverse of a transformed exponential function is always a transformed logarithmic function

How does this work in practice?

• E.g. to find the inverse of $f\left(x\right)=3·2^{(x+1)}-5$

• Think about the operations that transform $x$ into $f\left(x\right)$

  • For $f\left(x\right)=3·2^{(x+1)}-5$, the operations in order are

    • Add 1

      • $x\to x+1$

    • Raise 2 to the power of the result

      • $x+1\to 2^{\left(x+1\right)}$

    • Multiply by 3

      • $2^{\left(x+1\right)}\to 3·2^{\left(x+1\right)}$

    • Subtract 5

      • $3·2^{\left(x+1\right)}\to 3·2^{\left(x+1\right)}-5$

• To find the inverse, apply the inverse operations in the opposite order to $y=3·2^{(x+1)}-5$

  • Add 5 (reverses 'subtract 5')

    • $y+5=3·2^{(x+1)}$

  • Divide by 3 (reverses 'multiply by 3')

    • $\frac{y+5}{3}=2^{(x+1)}$

  • Take ${\log }_2$ (reverses 'raise 2 to the power')

    • and ${\log }_2$ cancels $2^{\square }$, so

    • ${\log }_2\left(\frac{y+5}{3}\right)=x+1$

  • Subtract 1 (reverses 'add 1')

    • ${\log }_2\left(\frac{y+5}{3}\right)-1=x$

  • Swap $x$ and $y$

    • $f^{-1}\left(x\right)={\log }_2\left(\frac{x+5}{3}\right)-1$

What is the domain of the inverse?

• The domain of the inverse function equals the range of the original function

  • and vice versa

• E.g. for $f\left(x\right)=a{b}^{(x+h)}+k$

  • If $a>0$

    • the range of $f$ is $(k,\infty )$

      • so the domain of $f^{-1}$ is $(k,\infty )$

  • If $a<0$

    • the range of $f$ is $(-\infty ,k)$, so the domain of $f^{-1}$ is $(-\infty ,k)$

• This makes sense because the logarithmic expression in the inverse $f^{-1}\left(x\right)={\log }_b\left(\frac{x-k}{a}\right)-h$ requires a positive argument

Inverses of logarithmic functions

How do I find the inverse of a transformed logarithmic function?

• A transformed logarithmic function has the general form $f\left(x\right)=a{\log }_b(x+h)+k$

• This is a combination of

  • additive transformations

    • horizontal shift $h$

    • vertical shift $k$

  • and a vertical dilation

    • factor $a$

  • applied to the base logarithmic function ${\log }_{b}x$

• As with finding the inverse of an exponential, find the inverse by reversing the operations in the opposite order

• In general (using $y=a{\log }_b(x+h)+k$)

  • Start with the equation

    • $y=a{\log }_b(x+h)+k$

  • Subtract $k$

    • $y-k=a{\log }_b(x+h)$

  • Divide by $a$

    • $\frac{y-k}{a}={\log }_b(x+h)$

  • Convert to exponential form ($b^{\square }$ cancels ${\log }_b$ on the right-hand side)

    • $b^{(y-k)/a}=x+h$

  • Subtract $h$

    • $b^{(y-k)/a}-h=x$

  • Swap $x$ and $y$:

    • $f^{-1}\left(x\right)=b^{(x-k)/a}-h$

• The inverse of a transformed logarithmic function is always a transformed exponential function

• E.g. to find the inverse of $f\left(x\right)=2{\log }_3(x-4)+1$

  • follow the steps given above

    • $y=2{\log }_3(x-4)+1$

    • $y-1=2{\log }_3(x-4)$

    • $\frac{y-1}{2}={\log }_3(x-4)$

    • $3^{(y-1)/2}=x-4$

    • $x=3^{(y-1)/2}+4$

    • $f^{-1}\left(x\right)=3^{(x-1)/2}+4$

What is the domain of the inverse?

• The domain of the inverse function equals the range of the original function

  • and vice versa

• For $f\left(x\right)=a{\log }_b(x+h)+k$

  • The range of a logarithmic function (in general form) is all real numbers

  • So the domain of the inverse (which is an exponential function) is also all real numbers

• This is consistent with what you know about exponential functions having a domain of all real numbers