solve the equation.

Question

log2(x+4)-log4x=2

please show work

Answer 1

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