How do you solve ${log}_{2} (2 x) = - 0.65$ $log_2(2x)=-0.65$ ?

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Answer 1 · 2018-03-06T22:10:04

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

By the law of logarithms:

If:

$y = {log}_{b} a$ $y=log_ba$

Then:

$b^{y} = a$ $b^y=a$

Since: $y = {log}_{b} a$ $y=log_ba$

$b^{{log}_{b} a} = a$ $b^(log_ba)=a$

Using this idea:

${log}_{2} (2 x) = - 0.65$ $log_2(2x)=-0.65$

$2^{{log}_{2} (2 x)} = 2^{- 0.65}$ $2^(log_2(2x))=2^(-0.65)$

$2 x = 2^{- 0.65}$ $2x=2^(-0.65)$

Dividing by $2$ $2$ :

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

How do you solve log2(2x)=−0.65log_2(2x)=-0.65?