Answer
$log_2 ~5$ is an irrational number.
Work Step by Step
Let's assume that $log_2 ~5$ is a rational number.
Then:
$log_2~5 = \frac{a}{b}$ (where $a$ and $b$ are positive integers)
$2^{a/b} = 5$
$(2^a)^{1/b} = 5$
$2^a = 5^b$
However, $2^a$ is an even number while $5^b$ is an odd number. The statement $2^a=5^b$ is a contradiction.
Therefore, our initial assumption must be false. Thus, $log_2 ~5$ must be an irrational number.
