Question

what is the inverse of the function?
$f(x) = 3^x - 1$
$\circ$ $f^{-1}(x) = \log_3(x) + 1$
$\circ$ $f^{-1}(x) = \log_3(x - 1)$
$\circ$ $f^{-1}(x) = \log_3(x + 1)$
$\circ$ $f^{-1}(x) = \log_3(x) - 1$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x) = 3^x - 1 \). Replace \( f(x) \) with \( y \), so we have \( y = 3^x - 1 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = 3^y - 1 \).

Step3: Solve for \( y \)

First, add 1 to both sides of the equation: \( x + 1 = 3^y \).
Then, we use the definition of a logarithm. Recall that if \( a^b = c \), then \( \log_a(c) = b \). Here, \( a = 3 \), \( b = y \), and \( c = x + 1 \). So we can rewrite \( 3^y = x + 1 \) as \( y = \log_3(x + 1) \).

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

Now, replace \( y \) with \( f^{-1}(x) \) to get the inverse function: \( f^{-1}(x) = \log_3(x + 1) \).

Answer:

\( f^{-1}(x)=\log_{3}(x + 1) \) (the third option: \( f^{-1}(x)=\log_{3}(x + 1) \))