How do you solve $- 5 = {log}_{2} x$ $-5 = log_2 x$ ?

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How do you solve $- 5 = {log}_{2} x$ $-5 = log_2 x$ ?

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Answer 1 · 2015-10-12T09:49:23

$x = \frac{1}{32}$ $x=1/32$

Explanation:

${log}_{2} x = - 5$ $log_2 x=-5$

Using the rule ${log}_{a} M = N \Rightarrow M = a^{N}$ $log_a M=N rArr M=a^N$ , you can rewrite ${log}_{2} x = - 5$ $log_2x=-5$ as $x = 2^{- 5}$ $x=2^-5$

Using the indices rule $a^{- m} = \frac{1}{a^{m}}$ $a^-m=1/a^m$ ,

$x = \frac{1}{2^{5}}$ $x=1/2^5$

$x = \frac{1}{32}$ $x=1/32$

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How do you solve $- 5 = {log}_{2} x$ $-5 = log_2 x$ ?

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