We have f: R → (–1, 1) and

Given that f^-1 exists.

Let y ϵ (–1, 1) such that f(x) = y

⇒ 10^2x – 1 = y (10^2x + 1)

⇒ 10^2x – 1 = 10^2xy + y

⇒ 10^2x – 10^2xy = 1 + y

⇒ 10^2x (1 – y) = 1 + y

Taking log₁₀ on both sides, we get

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus,