Question

4x find the inverse of the function $f(x) = \log_{8}(x - 1)$. 4x a $f^{-1}(x) = 8^{x + 1}$ 4x b $f^{-1}(x) = 8^{x} + 1$ 4x c $f^{-1}(x) = 8^{x}$ 4x d $f^{-1}(x) = x^{8 - 1}$

Explanation:

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

We start by writing the function as \( y = \log_{8}(x - 1) \).

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

To find the inverse, we interchange \( x \) and \( y \), so we get \( x = \log_{8}(y - 1) \).

Step3: Convert to exponential form

Recall that \( \log_{b}(a)=c \) is equivalent to \( b^{c}=a \). Applying this, we have \( 8^{x}=y - 1 \).

Step4: Solve for \( y \)

Add 1 to both sides of the equation: \( y = 8^{x}+1 \). Then we replace \( y \) with \( f^{-1}(x) \), so \( f^{-1}(x)=8^{x}+1 \).

Answer:

B. \( f^{-1}(x)=8^{x}+1 \)