Question

what is the inverse of the function?
f(x) = 3^x - 1
○ f⁻¹(x) = log₃(x) + 1
○ f⁻¹(x) = log₃(x - 1)
○ f⁻¹(x) = log₃(x + 1)
○ f⁻¹(x) = log₃(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, isolate the exponential term. Add 1 to both sides of the equation: \( x + 1 = 3^y \).

Now, we need to solve for \( y \) in terms of \( x \). Since the exponential function \( 3^y \) and the logarithmic function \( \log_3(z) \) are inverses of each other (i.e., if \( z = 3^y \), then \( y = \log_3(z) \)), we can rewrite \( x + 1 = 3^y \) in logarithmic form.

Applying the definition of logarithms, if \( a^b = c \), then \( b = \log_a(c) \). Here, \( a = 3 \), \( b = y \), and \( c = x + 1 \). So we get \( y = \log_3(x + 1) \).

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

Now that we have solved for \( y \) in terms of \( x \), we replace \( y \) with \( f^{-1}(x) \) to denote the inverse function. So \( f^{-1}(x) = \log_3(x + 1) \).

Answer:

\( f^{-1}(x) = \log_3(x + 1) \) (corresponding to the option \( f^{-1}(x)=\log_{3}(x + 1) \))