Answer
$x=\displaystyle \frac{2}{5}$
Work Step by Step
First, the solutions must satisfy $\left\{\begin{array}{l}
x+6\gt 0\\
x \gt 0
\end{array}\right.\quad \Rightarrow x\gt 0\qquad (*)$
in order for the equation to be defined.
LHS: Apply$ \displaystyle \quad\log_{a}\frac{M}{N}=\log_{a}M-\log_{a}N$
RHS: Apply $\quad \log_{4}4^{2}=2$
$\displaystyle \log_{4}(\frac{x+6}{x})=\log_{4}16$
... apply the principle of logarithmic equality
$\displaystyle \frac{x+6}{x}=16$
$x+6=16x$
$6=15x$
$x=\displaystyle \frac{6}{15}=\frac{2}{5}\qquad $... satisfies (*), and is a valid solution.
