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 $a<1$.)