Find the inverse of the function f(x) = \(log_a(x + \sqrt{(x^2+1)})\); a > 1 and assuming it to be an onto function.

Solution :

Given f(x) = \(log_a(x + \sqrt{(x^2+1)})\)

f'(x) = \(log_ae\over {\sqrt{1+x^2}}\) > 0

which is strictly increasing functions.

Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible.

Interchanging x & y

\(\implies\) \(log_a(y + \sqrt{(y^2+1)})\) = x

\(\implies\) \(y + \sqrt{(y^2+1)}\) = \(a^x\) ……..(1)

and \(\sqrt{(y^2+1)}\) – y = \(a^{-x}\) ………..(2)

From (1) and (2), we get y = \(1\over 2\)(\(a^x – a^{-x}\)) or \(f{-1}\)(x) = \(1\over 2\)(\(a^x – a^{-x}\)).