Answer
Solution set = $\{e^{2}-3\}.$
$ x\approx 4.39$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$ x+3 \gt 0$
$ x \gt -3 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\ln\sqrt{x+3}=1\qquad $... apply $ e^{(...)}$, the inverse of $\ln $
$e^{\ln\sqrt{x+3}}=e^{1}$
$\sqrt{x+3}=e\qquad $... square both sides
$ x+3=e^{2}\qquad $... subtract 3
$ x=e^{2}-3\qquad $... round to 2 decimal places
$ x\approx4.39$
(satisfies the condition (*), and is thus a valid solution.)