# math

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solve the equation.

log2(x+4)-log4x=2

• math -

we should have the same base in the logs

let log4x = y
then 4^y = x
2^(2y) = x
log2 x = 2y
y = (1/2)log2 x
= log2 x^(1/2)
= log2√x

then log4x = log2 √x

log2[ (x+4)/√x] = 2

(x+4)/√x = 2^2 = 4
x+4 = 4√x
square both sides
x^2 + 8x + 16 = 16x
x^2 - 8x + 16 = 0
(x-4)^2 = 0
x-4 = 0

x = 4

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