how to solve this question
$\displaystyle
2+\log_2{(3x-1)}=\log_2{4x}
$
1. Re-write the equation:
$\displaystyle 2+\log_2{(3x-1)}=\log_2{4x}~\implies~2=\log_2{4x} - \log_2{(3x-1)}~\implies~ $$\displaystyle 2=\log_2\left( \frac{4x}{(3x-1)}\right)$
2. Now use the base 2 on both sides of the equation:
$\displaystyle 4=\frac{4x}{(3x-1)}$ Solve for x. Keep in mind that $\displaystyle x > \tfrac13$.