Question

find $f^{-1}(x)$ given that $f(x) = 8^x$.
$f^{-1}(x) = $

Explanation:

Step1: Let \( y = f(x) \)

Set \( y = 8^x \). This is the original function with \( y \) representing the output of \( f(x) \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \), so we get \( x = 8^y \).

Step3: Solve for \( y \) using logarithms

We know that if \( a^b = c \), then \( b=\log_a c \). Here, \( a = 8 \), \( b = y \), and \( c=x \), so \( y=\log_8 x \). By the definition of an inverse function, \( f^{-1}(x)=\log_8 x \) (we can also write this using the change - of - base formula as \( \frac{\ln x}{\ln 8} \) or \( \frac{\log x}{\log 8} \), but \( \log_8 x \) is the most direct form for the inverse of an exponential function with base 8).

Answer:

\(\log_{8}x\) (or \(\frac{\ln x}{\ln 8}\) or \(\frac{\log x}{\log 8}\))