Prove that the following function $f(x)=\log_{2}\left(x+{\sqrt{x^2+1}}\right)$ is invertible on the whole number line.

Question

I think that it would help to show that this function is odd. But how can I show that the function $$\log_{2}\left(x+{\sqrt{x^2+1}}\right)$$ is invertible?

Answer 1

You could also directly use that an explicit formula for the Area sinus hyperbolicus is $$ \text{Arsinh}(x)=\ln(x+\sqrt{x^2+1}) $$ since $$ x=\sinh(y)=(e^y-e^{-y})/2 $$ can be written as quadratic equation $$ 0=(e^y)^2-2e^yx-1=(e^y-x)^2-(1+x^2) $$ with the only positive solution $$ e^y=x+\sqrt{x^2-1}\implies y=\ln(x+\sqrt{x^2+1}). $$

Answer 2

You just have to prove it is strictly monotonic. Can you determine the sign of its derivative?

Answer 3

Check that $x+\sqrt {x^2+1}$ maps $\mathbb {R}$ bijectively onto $(0,\infty).$ Then recall $\log_2 x$ maps $(0,\infty)$ bijectively onto $\mathbb {R}.$ It follows that the composition is a bijection of $\mathbb {R}$ onto $\mathbb {R},$ which gives the conclusion.