Respuesta :

arthurpdc
Firstly, we'll find y = f⁻¹(x):

[tex]f(x)=\log_3(x+1)\\\\ x=\log_3(y+1)\\\\ y+1=3^x\\\\ y=3^x-1\\\\ f^{-1}(x)=3^x-1 [/tex]

So:

[tex]f^{-1}(x)=3^x-1\\\\ f^{-1}(2)=3^2-1\\\\ f^{-1}(2)=9-1\\\\ \boxed{f^{-1}(2)=8}[/tex]

Answer: The value is 8.

Step-by-step explanation:  Given function is

[tex]f(x)=\log_3(x+1).[/tex]

we are given to find the value of [tex]f^{-1}(2).[/tex]

Let us consider that

[tex]f(x)=y~~~~~\Rightarrow x=f^{-1}(y).[/tex]

So, we have

[tex]f(x)=\log_3(x+1)\\\\\Rightarrow y=\log_3(f^{-1}(y)+1)\\\\\Rightarrow 3^y=f^{-1}(y)+1\\\\\Rightarrow f^{-1}(y)=3^y-1.[/tex]

If y = 2, then we have

[tex]f^{-1}(2)=3^2-1=9-1=8.[/tex]

Thus, the required value is 8.