Answer
Solution set = $\displaystyle \{\frac{e^{4}}{2}\}.$
$ x\approx 27.30$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$2x \gt 0$
$ x \gt 0 \qquad (*)$
This is the condition eventual solutions must satisfy.
$ 5\ln 2x=20\qquad $... divide with $5$
$\ln 2x=4\qquad $... apply $ e^{(...)}$, the inverse of $\ln $
$ e^{\ln 2x}=e^{4}$
$ 2x=e^{4}\qquad $... divide with $2$
$ x=\displaystyle \frac{e^{4}}{2}\qquad $... round to 2 decimal places
$ x\approx 27.30$
(satisfies the condition (*), and is thus a valid solution.)