log base (2) (x) +long base(2) (3) = log base(2) (18)
You can use the rules for calculating with logarithms. In general
$\displaystyle \log_a x + \log_a y = \log_a (xy)$
and
$\displaystyle \log_a x - \log_a y = \log_a \left(\frac{x}{y}\right)$.
For your problem
$\displaystyle \log_2 x - \log_2 18 + \log_2 3 = 0$
transforms to
$\displaystyle \log_2\left(\frac{3x}{18}\right) = 0$
or
$\displaystyle \frac{3x}{18} = 1$