Question

If f(x) = log2 (x + 1), what is f-1(2)? (2 points)

Answer 1

Answer:

3

Step-by-step explanation:

$f^{-1}(2)=b \text{ implies } f(b)=2$

This we means to to solve the following equation for b:

$f(b)=\log_2(b+1)$

$2=\log_2(b+1)$ since f(b)=2

Write in equivalent exponential form:

$2^2=b+1$

$4=b+1$

Subtract 1 on both sides:

$4-1=b$

$3=b$

This means $f(3)=2 \text{ so } f^{-1}(2)=3$

You could actually find the inverse function if you want to then replace input for the inverse with 2.

$y=\log_2(x+1)$

Your logarithm has base 2 and input (x+1) and output y.

The equivalent exponential form is:

$2^{y}=x+1$

If we solve for x then at the end swap x and y we would have found the inverse function.

Let's do that:

$2^y=x+1Subtract 1 on both sides:[tex]2^y-1=x$

Swap x and y:

$2^x-1=y$

The inverse function of our given function is:

$f^{-1}(x)=2^x-1$

Now we need to replace x with 2:

$f^{-1}(2)=2^2-1$

$f^{-1}(2)=4-1$

$f^{-1}(2)=3$

Answer 2

as you already know, to get the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".

$\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=\log_2(x+1)\implies \stackrel{\textit{quick switcheroo}}{\underline{x}=\log_2(\underline{y}+1)}\implies 2^x=2^{\log_2({y}+1)} \\\\\\ 2^x=y+1\implies 2^x-1=\stackrel{f^{-1}(x)}{y} \\\\[-0.35em] ~\dotfill\\\\ 2^2-1=f^{-1}(2)\implies 3=f^{-1}(2)$