Question #b370e

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Answer 1 · 2017-11-03T14:55:26

x=4

$log_2 x + log_2 (x+4) = 5$

$log_2 (x * (x+4)) = 5$

$log_2 (x^2 + 4x) = 5$

$x^2 + 4x = 2^5 = 32$

i.e.

$x^2 +4x - 32 = 0$

$x1 = (-4 + sqrt(4^2 - 4*1*(-32)))/(2*1)$
and
$x2 = (-4 - sqrt(4^2 - 4*1*(-32)))/(2*1)$

which are

$x1 = (-4 + sqrt(16 +128))/2$
and
$x2 = (-4 - sqrt(16+128))/2$

which becomes

$x1 = -2 + 6 = 4$
and
$x2 = -2 - 6 = -8$

Check if true:

$log_2 (4) + log_2 (4+4) = 2 + 3 = 5$ , so x1 is correct,
and
$log_2(-8) +...$ , nope, you cannot take the logarithm of a negative number.

So the only answer is x= 4.
Q.E.D.