How do you solve ${log}_{2} (2 x) + {log}_{2} x = 5$ $log_2( 2x) + log _ 2 x = 5$ ?

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How do you solve ${log}_{2} (2 x) + {log}_{2} x = 5$ $log_2( 2x) + log _ 2 x = 5$ ?

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Answer 1 · 2015-11-27T00:18:05

I found $x = 4$ $x=4$

Explanation:

We can use the property of logs relating the sum with the product of the arguments and write:
${log}_{2} (2 x \cdot x) = 5$ $log_2(2x*x)=5$
use the definition of log:
$2 x^{2} = 2^{5}$ $2x^2=2^5$
$x^{2} = \frac{32}{2}$ $x^2=32/2$
$x = \pm \sqrt{16} = \pm 4$ $x=+-sqrt(16)=+-4$
only the positive one is acceptable.

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How do you solve ${log}_{2} (2 x) + {log}_{2} x = 5$ $log_2( 2x) + log _ 2 x = 5$ ?

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Explanation:

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How do you solve ${log}_{2} (2 x) + {log}_{2} x = 5$ $log_2( 2x) + log _ 2 x = 5$ ?