Simplify $\sinh (\log (x))$

Question

$$\sinh (\log (x))=\frac{x^2-1}{2 x}$$

However I do not see how this is done, here is an idea I had but I'm probably way off:

$$\sinh \left(\ln \left(\frac{1}{2} \left(e^x-e^{-x}\right)\right)\right)$$

Here I can take out the constant but I do not understand what I am allowed to do with $\text{Ln}\left(e^x-e^{-x}\right)$.

I am probably way off so and definitely far from completing it so could someone please help?

Answer 1

We have $$ \sinh(t) = \frac{e^{t}-e^{-t}}{2} = \frac{e^{2t}-1}{2e^t} \quad \text{ and } \quad e^{\ln(x)} = x.$$ So just replace $t$ by $\ln(x)$ to find your expression.

Answer 2

However I do not see how this is done.

That's because your second expression is $~\sinh\ln\sinh x~$ as opposed to $~\sinh\ln x$.