Proving $2\cosh 2x+ \sinh x = 5$

Question

I have been sitting on this question for quite some time and I haven't been able to prove this identity. Please anybody who can help me here. I am new to hyperbolics.

$$2\cosh 2x+ \sinh x = 5$$

I got the answer $\ln(2)$. But I don't want to solve the question is how to prove.

Answer 1

If $e^x=a$,

$$10=2(a^2+a^{-2})+a-a^{-1}=2(y^2+2)+y,$$ where $y=a-a^{-1}$

Can you take it from here?

Answer 2

Write $s:=\sinh x$ so $$0=2(2s^2+1)+s-5=4s^2+s-3=(s+1)(4s-3)\implies s\in\{-1,\,\tfrac34\}.$$From $\operatorname{arsinh}s=\ln(s+\sqrt{s^2+1})$, these values of $s$ give $x\in\{\ln(\sqrt{2}-1),\,\ln 2\}$.