# How do you solve log_2(3x-5)>log_2(x+7) and check the solutions?

Oct 25, 2016

$x > 6$

#### Explanation:

${\log}_{2} x$ is a monotonically increasing function of $x$.

Hence, if ${\log}_{2} \left(3 x - 5\right) > {\log}_{2} \left(x + 7\right)$

we have $3 x - 5 > x + 7$

or $3 x - x > 7 + 5$

or $2 x > 12$

or $x > 6$
graph{lnx/ln2 [-0.17, 19.83, -4.04, 5.96]}