Question

find $f^{-1}(x)$ given that $f(x) = 3^x$.
$f^{-1}(x) = $

Explanation:

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

We start by setting \( y = 3^x \). This is the original function, and we want to find its inverse, so we need to solve for \( x \) in terms of \( y \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). So we get \( x = 3^y \).

Step3: Solve for \( y \) using logarithms

We know that if \( a^b = c \), then \( b=\log_a c \). Here, \( a = 3 \), \( c=x \), and the exponent is \( y \). So we can rewrite \( x = 3^y \) in logarithmic form as \( y=\log_3 x \).

Step4: Replace \( y \) with \( f^{-1}(x) \)

Since we found \( y \) in terms of \( x \) and \( y = f^{-1}(x) \), we have \( f^{-1}(x)=\log_3 x \).

Answer:

\( \log_{3} x \) (or equivalently, using the change - of - base formula, it can also be written as \( \frac{\ln x}{\ln 3} \) or \( \frac{\log x}{\log 3} \), but the most direct form is \( \log_{3} x \))