If log(x-3)+log(x+4)-log x=log 5 then the solution set for x is?:

a.{2,6}
b.{-2,6}
c.{6}
d.{-6,2}

log(x-3)(x+4)-logx=log5
logx^2+4x-3x-12-logx=log5
logx^2+x-12/x=log5
(x^2+x-12/x)=5
x^2+x-12=5x
x^2-4x-12=0
(x-6)(x+2)=0
x= -2, 6
Make sure both solutions work!
Answer: C

[x^2+x--12]/x=5

x^2+x--12--5x=0

x^2-4x--12=0

x^2--6x+2x--12=0

x[x--6]+2[x--6]=0

[x--6][x+2]=0

x=6 0r x=--2

Hence option b (--2,6) is the correct solution set

( x - 3 ) ( x + 4 ) / x = 5

x ² + x - 12 = 5x

x ² - 4x - 12 = 0

( x - 6 ) ( x + 2 ) = 0

x = 6

OPTION C