How do you solve $log_2(x+2) - log_2x = 3$ ?

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Answer 1 · 2016-03-31T06:39:16

$x = 2/7$

Using log a - log b = $log(a/b)$ ,
$log_2((x+2)/x) = 3$
Inversely, $((x+2)/x)=2^3=8$
7x = 2.

How do you solve log_2(x+2) - log_2x = 3log_2(x+2) - log_2x = 3?