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