2007-11-05 06:09:21 · 4 answers · asked by malibu0108 1 in Science & Mathematics ➔ Mathematics
log(a) - log(b) = log(a/b)
log(3+x) - log(4-x) = log[(3+x)/(x-4)] = log(2)
(3+x) / (x-4) = 2
3 + x = 2x - 8
x = 11
2007-11-05 06:17:40 · answer #1 · answered by Helen B 5 · 0⤊ 0⤋
log(3+x) = log2 + log(x-4) = log[2(x-4)]
3+x = 2(x-4) = 2x - 8
Solve for x,
x = 11
2007-11-05 06:13:15 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
log(3+x)-log(x-4) = log2
log(3+x) = log2+log(x-4)
log(3+x) = log[2(x-4)]
10^log(3+x) = 10^[log(2(x-4))]
3+x=2(x-4)
3+x=2x-8
3+8=2x-x
11=x
2007-11-05 06:27:40 · answer #3 · answered by Anonymous · 0⤊ 0⤋
If you place inverse log on each side you will get
(3+x)/(x-4) = 2
multiply both sides by (x-4)
3+x = 2x-8
solve for x
x = 11
2007-11-05 06:13:50 · answer #4 · answered by Ilya S 3 · 0⤊ 0⤋