How to show that $x=\ln 3$  solves $x=\ln(10/3 - e^{-x})$ My homework has tasked me with finding $x$ when $\cosh x=5/3$. I know that the solution is $\ln (3)$, but I can't figure out how to solve it myself. The furthest I can simplify it is the following:
$$\frac{e^x+e^{-x}}{2} = 5/3$$
$$ e^x+e^{-x} = 10/3$$
$$e^x = \frac{10}{3}-e^{-x}$$
$$x = \ln \left(\frac{10}{3}-e^{-x} \right)$$
Now, if I put this into Wolfram Alpha it tells me that the answer is $\ln(3)$, but it doesn't tell me how it solved that. Also, I'm guessing there may be another way to solve this by taking a different route than the above. 

Start with $$e^x+e^{-x}={10\over3}$$
Multiply both sides by $e^x$:
$$
e^x\cdot e^x+e^x\cdot e^{-x}={10\over 3}e^x
$$
Simplify:
$$
(e^x)^2+1={10\over 3}e^x.
$$
Let $u=e^x$, then
$$
u^2+1={10\over3}u
$$
or
$$
3u^2-10u+3=0.
$$
This has solutions $u=3$ and $u=1/3$.
So $e^x=3$ or $e^x=1/3$.
