## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$16$
The domain of the variable requires that $x>0$. Recall: $\log_a M^r = r \log_a M$ This means that: $\dfrac{1}{2} \log_3 x = \log_3 x^{\frac{1}{2}}$ $2 \log_3 2 = \log_3 2^2$ Thus, the given equation is equivalent to: $\log_3 x^{\frac{1}{2}} = \log_3 2^2$ Recall also that: $\text{If } \log_a M = \log_a N \text{, then } M=N$ Therefore, $x^{\frac{1}{2}} = 2^2$ $x^{\frac{1}{2}} = 4$ $x=4^2$ $x= \boxed{16}$