2007-11-15 03:41:57 · 3 answers · asked by Smiles6 1 in Science & Mathematics ➔ Mathematics
log(x+3) + log(x-6) = 1
log((x+3)(x-6)) = log 10
(x+3)(x-6) = 10
x^2 - 3x - 18 = 10
x^2 - 3x -28 = 0
(x-7)(x+4) = 0
x = 7 or -4
x cannot be -4 because the arg of log cannot be negative
2007-11-15 03:53:02 · answer #1 · answered by norman 7 · 0⤊ 0⤋
log(x+3)(x-6)=1
x^2-3x-18=10
x^2-3x-28=0
(x-7)(x+4)=0
x=-4,7
however xcannot be -4 as log (-4+3) does not exist
x=7is the answer
2007-11-15 11:49:06 · answer #2 · answered by someone else 7 · 0⤊ 0⤋
log(a)+log(b)=log(ab)
log[ (x+3)(x-6)] = 0 ( log(1) = 0)
(x+3)(x-6)=10^0=1
we assume this is log (to base 10)
x^2-3x-18-1=0
x^2-3x-19=0
ax^2+bx+c=0
a=1 b=-3 c=-19
solve this quadratic equation for x
x=6.109, -3.109
2007-11-15 11:51:11 · answer #3 · answered by cidyah 7 · 0⤊ 0⤋