<s>[INST] How does $a^y = x+\sqrt{x^2+1}$ imply $a^{-y} = -x+ \sqrt{x^2+1}$ and then $x=\sinh(y\ln a)$? 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 $(3)$.
 [/INST] 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 $x = \frac{1}{2}(a^y-a^{-y})$.
</s>