Answer
Solution set = $\{e^{-1/3}\}.$
$ x\approx 0.72$
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.
$ 7+3\ln x=6\qquad $... subtract 7
$ 3\ln x=-1\qquad $... divide with 3
$\displaystyle \ln x=-\frac{1}{3}\qquad $... apply $ e^{(...)}$, the inverse of $\ln $
$ e^{\ln x}=e^{-1/3}$
$ x=e^{-1/3}\qquad $... round to 2 decimal places
$ x\approx 0.72$
(satisfies the condition (*), and is thus a valid solution.)