Answer
$5$
Work Step by Step
The domain of the variable requires that $x+4>0$.
This means that $x>-4$.
Recall:
$\log_a M^r = r \log_a M$
Using the rule above gives:
$2 \log_3 (x+4) = \log_3 (x+4)^2$
Thus, the given equation is equivalent to:
$\log_3 (x+4)^2 - \log_3 9 = 2$
Recall further that:
$\log_a\left(\dfrac{M}{N}\right) = \log_a M-\log_a N$
Hence, the equation above is equivalent to:
$\log_3 \left(\dfrac{(x+4)^2}{9} \right) = 2$
Since $y = \log_a b \text{ is equivalent to } b=a^y$, then
$\log_3 \left(\dfrac{(x+4)^2}{9} \right) = 2$ is equivalent to $\dfrac{(x+4)^2}{9} = 3^2$.
Solve the equation to obtain:
$\dfrac{(x+4)^2}{9} = 9$
$(x+4)^2 = 9(9)$
$(x+4)^2 = 81$
$x+4 = \pm \sqrt{81}$
$x+4 = \pm 9$
Thus,
$x+4 = 9 \hspace{15pt} \to x = 5$
$x+4 = -9 \hspace{15pt} \to x=-13 \hspace{5pt} \text{(Rejected since $x$ must be greater than $-4$)}$
Therefore,
$x = \boxed{5}$
