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.