Given the function f(x) = log5(x + 1), find the value of f^-1(2).

Solution:

Given the function f(x) = log5(x + 1)

A Function in math is visualized as a rule, which gives a unique output for every input x.

The logarithm of any number N if interpreted as an exponential form, is the exponent to which the base of the logarithm should be raised, to obtain the number N. 

Let y = f(x) = log5(x + 1),

e^y = 5(x + 1)

e^y / 5 = (x + 1)

(e^y / 5) - 1 = x

y = f(x)

=> x = f^-1(y)

f^-1(2) = e² / 5 - 1

f^-1(2) = -0.55

Therefore, the value of f^-1(2) for the function f(x) = log5(x + 1) is -0.55

Given the function f(x) = log5(x + 1), find the value of f^-1(2).

Summary:

Given the function f(x) = log5(x + 1),the value of f^-1(2) is -0.55.