Answer
Solution set = $\{e^{2}-4\}.$
$ x\approx 3.39$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$ x+4 \gt 0$
$ x \gt -4 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\ln\sqrt{x+4}=1\qquad $... apply $ e^{(...)}$, the inverse of $\ln $
$ e^{\ln\sqrt{x+4}}=e^{1}$
$\sqrt{x+4}=e\qquad $... square both sides
$ x+4=e^{2}\qquad $... subtract $4$
$ x=e^{2}-4\qquad $... round to 2 decimal places
$ x\approx 3.39.$
(satisfies the condition (*), and is thus a valid solution.)