hey y'all,

I got a question (obviously)

"By expressing $\displaystyle y = \sinh{(x)}$ as a quadratic in the term $\displaystyle e^x$, show $\displaystyle \sinh^{-1}(x) = \ln{(x + \sqrt{x^2 +1})}$."

This is my worked solution for the quadratic part:

$\displaystyle y = sinh(x) = \frac{1}{2}(e^x - e^{-x})$

$\displaystyle y * 2e^x = {[\frac{1}{2}(e^x - e^{-x}]}*2e^x = {(e^x)}^2 - 1$

$\displaystyle 0 = e^{2x} + 2ye^x - 1$

And now for the inverse part:

I get stuck i know the answer is $\displaystyle \sinh^{-1}(x) = \ln{(x + \sqrt{x^2 +1})}$ (from the question) and i know somewhere in the calculations there is some sort of logarithm cancellation i.e. $\displaystyle \ln{(e^x)} = x$

Can someone help? and can you tell me if my quadratic expression is incorrect? thanks