Question

If f(x) = log3 (x + 1), what is f−1(2)

Answer 1

Firstly, we'll find y = f⁻¹(x):

$f(x)=\log_3(x+1)\\\\ x=\log_3(y+1)\\\\ y+1=3^x\\\\ y=3^x-1\\\\ f^{-1}(x)=3^x-1$

So:

$f^{-1}(x)=3^x-1\\\\ f^{-1}(2)=3^2-1\\\\ f^{-1}(2)=9-1\\\\ \boxed{f^{-1}(2)=8}$

Answer 2

Answer: The value is 8.

Step-by-step explanation:  Given function is

$f(x)=\log_3(x+1).$

we are given to find the value of $f^{-1}(2).$

Let us consider that

$f(x)=y~~~~~\Rightarrow x=f^{-1}(y).$

So, we have

$f(x)=\log_3(x+1)\\\\\Rightarrow y=\log_3(f^{-1}(y)+1)\\\\\Rightarrow 3^y=f^{-1}(y)+1\\\\\Rightarrow f^{-1}(y)=3^y-1.$

If y = 2, then we have

$f^{-1}(2)=3^2-1=9-1=8.$

Thus, the required value is 8.