Question

log2 (2x) + log2 (3x+4)=6

Answer 1

log2 2x(3x+4) = log2 64
6x² + 8x = 64
6x² + 8x - 64 = 0
6x² + 24x - 16x - 64 =0
6x (x+4) -16(x+4) = 0
(6x - 16 )(x+4) = 0

6x = 16
x = 16/6
x = 2 2/3

x = -4

{2 2/3, -4}