How do you solve $15+2log_2x=31$ ?

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Answer 1 · 2016-09-07T14:58:02

$x = 256$

We have: $15 + 2 log_(2)(x) = 31$

First, let's subtract $15$ from both sides of the equation:

$=> 2 log_(2)(x) = 16$

Then, let's divide both sides by $2$ :

$=> log_(2)(x) = 8$

Now, using the laws of logarithms:

$=> x = 2^(8)$

$=> x = 256$

Therefore, the solution to the equation is $x = 256$ .

Answer 2 · 2016-09-07T15:00:04

$x=256$

$15+2log_2 x=31$

$hArr2log_2 x=31-15=16$ or

$log_2=8$

Hence $x=2^8=256$

How do you solve 15+2log_2x=3115+2log_2x=31?