How does $a^y = x+\sqrt{x^2+1}$ imply $a^{-y} = -x+ \sqrt{x^2+1}$ and then $x=\sinh(y\ln a)$?

Question

I want to understand the steps in an argument given for the following question.

Find the inverse of the following function: $$f(x) = \log_a(x+\sqrt{x^2+1})$$ We find: $$\begin{align} a^y &= \phantom{-}x+\sqrt{x^2+1} \tag1 \\[4pt] a^{-y} &= -x+ \sqrt{x^2+1} \tag2 \end{align}$$ whence: $$x = \frac{1}{2} (a^y-a^{-y}) = \sinh(y\ln a) \tag3$$

I understand how he converts it to an exponential $(1)$, but then $(2)$ doesn’t make sense to me, as well as part $(3)$.

Answer 1

Right, so what you should do is to write these things out explicitly. So: $$a^y = \sqrt{x^2+1}+x$$ $$a^{-y} = \frac{1}{a^y} = \frac{1}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}+x} \cdot \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x} = \frac{\sqrt{x^2+1}-x}{(x^2+1)-x^2} = \sqrt{x^2+1}-x$$ Then, notice that: $$a^y-a^{-y} = (\sqrt{x^2+1}+x)-(\sqrt{x^2+1}-x) = 2x$$ It follows that $x = \frac{1}{2}(a^y-a^{-y})$.