How do you solve $log_2 (4x)=5$ ?

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Answer 1 · 2016-07-11T13:12:09

8

in a variety of ways

for example

$log_2 (4x)=5$

$implies log_2 4 + log_2 x=5$
$implies 2 + log_2 x=5$
$implies log_2 x=3$
$implies x=2^3 = 8$

OR

$log_2 (4x)=5$
$2^(log_2 (4x))=2^5$
$4x=32$
$x = 8$

OR

simplest, maybe, taken straight from the idea that a logarithm is just an index

$log_2 (4x)=5$
$implies 4x=2^5$
$implies x=2^5* 2^(-2) = 2^3 = 8$

How do you solve log_2 (4x)=5log_2 (4x)=5?