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. 
 A: 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 $e^x=1/3$.
A: Here is something a little different that gives us a chance to play with hyperbolic functions.
Recall that
$$\cosh x=\frac{e^x+e^{-x}}{2} \qquad \text{and}\qquad \sinh x=\frac{e^x-e^{-x}}{2}.$$
The fact that $\cosh(-x)=\cosh x$ and $\sinh(-x)=-\sinh x$ can be verified directly from the definitions. 
The result for $\cosh x$ shows that if we find a solution $x$ of the equation $\cosh x=a$, then $-x$ is also a solution of the equation.
From the definitions of $\cosh x$ and $\sinh x$, it is not hard to verify that
$$\cosh^2 x-\sinh^2x=1.\qquad(\ast)$$
This identity, so reminiscent of $\cos^2 x+\sin^2 x=1$, is the key to the importance of $\cosh x$ and $\sinh x$, and to the computation that follows.
Enough of previews. It is time for the movie.

If $\cosh x =\frac{5}{3}$, then by $(\ast)$ $\sinh^2 x=\cosh^2 x-1=\frac{25}{9}-1=\frac{16}{9}$, and therefore $\sinh x=\pm \frac{4}{3}$. Thus
$$\frac{e^x+e^{-x}}{2}=\frac{5}{3} \qquad \text{and}\qquad \frac{e^x-e^{-x}}{2}=\pm \frac{4}{3}.$$
Add. We get
$$e^x=\frac{5}{3}\pm \frac{4}{3}.$$
Thus $e^x=\frac{9}{3}=3$ or $e^x=\frac{1}{3}$. It follows that $x=\ln 3$ or $x=\ln(1/3)=-\ln 3$.
Comment: Exactly the same method can be used to solve $\cosh x=a$. (There is no real solution if $a<1$.) 
A: as @David Mitra said you made a light misstep in the third step 
$$e^x+\frac1{e^x}=\frac {10}{3}$$
$$3e^{2x}+3={10} e^x$$ 
$$3({e^x})^2-{10} e^x+3=0$$
then its just solving $e^x$ like any other quadratic (don't forget to eliminate the negative solution (if any))
