How do you solve $log_2(2x)=-0.65$ ?

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Answer 1 · 2018-03-06T22:10:04

$x=0.319color(white)(888)$ 3 .d.p.

By the law of logarithms:

If:

$y=log_ba$

Then:

$b^y=a$

Since: $y=log_ba$

$b^(log_ba)=a$

Using this idea:

$log_2(2x)=-0.65$

$2^(log_2(2x))=2^(-0.65)$

$2x=2^(-0.65)$

Dividing by $2$ :

$x=2^(-0.65)/2=0.319color(white)(888)$ 3 .d.p.

How do you solve log_2(2x)=-0.65log_2(2x)=-0.65?