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

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Answer 1 · 2016-10-25T14:40:02

$x>6$

$log_2 x$ is a monotonically increasing function of $x$ .

Hence, if $log_2(3x-5)>log_2(x+7)$

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

or $3x-x>7+5$

or $2x>12$

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

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