math
posted by anonymous on .
solve the equation.
log2(x+4)log4x=2
please show work

we should have the same base in the logs
let log_{4}x = y
then 4^y = x
2^(2y) = x
log_{2} x = 2y
y = (1/2)log_{2} x
= log_{2} x^(1/2)
= log_{2}√x
then log_{4}x = log_{2} √x
log_{2}[ (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
(x4)^2 = 0
x4 = 0
x = 4