How do you solve for x in $log_2x=log_4(25)$ ?

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Answer 1 · 2018-05-01T20:48:05

$x=5$

Using the formal definiton of a logarithm, we can take the right hand side to be:

$log_4 25=a=>4^a = 25$

Since $4$ is simply $2^2$ , we have:

$2^(2a)=25$

Remember; $a$ is equal to $log_4 25$ , which is in turn equal to $log_2 x$ .

$2^(2log_2x)=25$

Taking the binary logarithm of both sides:

$2log_2x=log_2 25$

$=> log_2x = 1/2log_2 25$

Knowing that $alog_b c = log_b c^a$ , we reach the result we wished to get:

$log_2x = log_2 (25)^(1/2) = log_2 5 => color(red)(x=5)$

How do you solve for x in log_2x=log_4(25)log_2x=log_4(25)?