Find all the real solutions of the following equation: log_4X + log_4 (x+12)=3?

Answer 1

log is log to base 4 in the following:-
log x + log (x + 12) = 3
log [ (x) (x + 12) ] = 3
(x) (x + 12) = 4³
x² + 12x = 64
x² + 12x - 64 = 0
(x + 16) (x - 4) = 0
x = - 16 , x = 4
Accept + ve value of x = 4

Answer 2

log_4X + log_4 (x+12)= 3

log_4X + log_4 (x+12)= log_4 (64)

x(x + 12) = 64 => x ² + 12x - 64 = 0

x' = 4 (the other x" = - 16 rejected )

Answer 3

log_4X + log_4 (x+12)=3
log_4 x(x+12)=3
log_4 (x^2+12x)=3
(x^2+12x)=4^3
x^2+12x-64=0
(x+16)(x-4)=0
x=-16 (undefined for logs) or x=4
So x=4
log4 4=1
log4 (4+12) = log4 16=2
1+2 = 3