# 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) = $$1\over 2$$($$a^x – a^{-x}$$).

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