Question

solve the logarithm equation,
log[4]x+log4(x+7)=log[4](x+27)

Answer 1

is that log base 4 ?

Answer 2

\[\log_{4}x+\log_{4}(x+7)=\log_{4}(x+27) \]

Answer 3

log[4] x(x+7) = log[4] (x+27)
x^2+7x = x + 27
x^2 + 6x - 27 = 0
x^2 + 9x -3x -27
x(x+9)-3(x+9) = 0
(x+9)(x-3) = 0

x = -9, 3

but be ware!!
domain : x>0 , x+7>0 so x>-7, x+27>0 so x> -27 "intersection" between them all give you : x>0

so x=3.

Answer 4

this is the same as
\[\log_{4}(x^2+7x)=\log_{4}(x+27)\]because
log(a)+log(b)=log(ab)