Question

If e^y - e^-y= x, express y as an explicit function of x​

Answer 1

`e^y-e^(-y)=x\implies e^y-\cfrac{1}{e^y}=x\implies \cfrac{(e^y)^2-1}{e^y}=x \\\\\\ e^(2y)-1=xe^y \implies e^(2y)-xe^y=1`

now, we're going to do some "completing the square" on the left-side, that is, we'll make it a perfect square trinomial by using our very good friend Mr Zero, 0.

`2\cdot n\cdot e^y=xe^y\implies n=\cfrac{xe^y}{2e^y}\implies n=\cfrac{x}{2} \\\\[-0.35em] ~\dotfill`

`e^(2y)-xe^y=1\implies e^(2y)-xe^y + n^2 - n^2=1\implies e^(2y)-xe^y+n^2=1+n^2 \\\\\\ (e^y)^2-xe^y+\left( \cfrac{x}{2} \right)^2=1+\cfrac{x^2}{4}\implies \left( e^y -\cfrac{x}{2} \right)^2=1+\cfrac{x^2}{4} \\\\\\ e^y -\cfrac{x}{2}=\sqrt{1+\cfrac{x^2}{4}}\implies e^y =\sqrt{1+\cfrac{x^2}{4}}+\cfrac{x}{2} \\\\\\ \ln(e^y)= \ln\left(\cfrac{x}{2}+ \sqrt{1+\cfrac{x^2}{4}} \right)\implies {\Large \begin{array}{llll} y=\ln\left(\cfrac{x}{2}+ \sqrt{1+\cfrac{x^2}{4}} \right) \end{array}}`