Respuesta :
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".
[tex]\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)[/tex]
Answer:
3
Step-by-step explanation:
[tex]f^{-1}(2)=b \text{ implies } f(b)=2[/tex]
This we means to to solve the following equation for b:
[tex]f(b)=\log_2(b+1)[/tex]
[tex]2=\log_2(b+1)[/tex] since f(b)=2
Write in equivalent exponential form:
[tex]2^2=b+1[/tex]
[tex]4=b+1[/tex]
Subtract 1 on both sides:
[tex]4-1=b[/tex]
[tex]3=b[/tex]
This means [tex]f(3)=2 \text{ so } f^{-1}(2)=3[/tex]
You could actually find the inverse function if you want to then replace input for the inverse with 2.
[tex]y=\log_2(x+1)[/tex]
Your logarithm has base 2 and input (x+1) and output y.
The equivalent exponential form is:
[tex]2^{y}=x+1[/tex]
If we solve for x then at the end swap x and y we would have found the inverse function.
Let's do that:
[tex]2^y=x+1
Subtract 1 on both sides:
[tex]2^y-1=x[/tex]
Swap x and y:
[tex]2^x-1=y[/tex]
The inverse function of our given function is:
[tex]f^{-1}(x)=2^x-1[/tex]
Now we need to replace x with 2:
[tex]f^{-1}(2)=2^2-1[/tex]
[tex]f^{-1}(2)=4-1[/tex]
[tex]f^{-1}(2)=3[/tex]
