Answer
Solution set = $\{e^{-1/2}\}.$
$ x\approx 0.61$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$ x \gt 0 \qquad (*)$
This is the condition eventual solutions must satisfy.
$ 6+2\ln x=5\qquad $... subtract 6
$ 2\ln x=-1\qquad $... divide with 2
$\displaystyle \ln x=-\frac{1}{2}\qquad $... apply $ e^{(...)}$, the inverse of $\ln $
$ e^{\ln x}=e^{-1/2}$
$ x=e^{-1/2}\qquad $... round to 2 decimal places
$ x\approx 0.61$
(satisfies the condition (*), and is thus a valid solution.)