$\displaystyle (3)(2^x)(3^x) = 10$
Take the logs (base 10 will keep it simpler) $\displaystyle log(3 \times 2^x \times 3^x) = 1$
Use the law $\displaystyle log_b(ac) = log_b(a) + log_b(c)$ to separate the left hand side
$\displaystyle log(3) + log(2^x) + log(3^x) = 1$
Take log(3) from both sides and then use the law which says $\displaystyle log(a^k) = k\, log(a)$
$\displaystyle x\, log(2) + x\, log(3) = 1-log(3)$
Factor out x: $\displaystyle x(log(2)+log(3)) = 1-log(3)$
Divide through to isolate x: $\displaystyle \frac{1-log(3)}{log(2)+log(3)}$