Answer
Solution set = $\displaystyle \{\frac{e^{5}}{2}\}.$
$ x\approx 74.21$
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.
$ 6\ln 2x=30\qquad $... divide with $6$
$\ln 2x=5\qquad $... apply $ e^{(...)}$, the inverse of $\ln $
$ e^{\ln 2x}=e^{5}$
$ 2x=e^{5}\qquad $... divide with $2$
$ x=\displaystyle \frac{e^{5}}{2}\qquad $... round to 2 decimal places
$ x=74.21$
(satisfies the condition (*), and is thus a valid solution.)