the bases are 2
2007-11-30 03:18:18 · 3 answers · asked by mouser 1 in Science & Mathematics ➔ Mathematics
log2(3x - 2) - log2(x - 5) = 4
log2 [(3x - 2) / (x - 5)] = 4
(3x - 2) / (x - 5) = 2^4
(3x - 2) / (x - 5) = 16
3x - 2 = 16(x - 5)
3x - 2 = 16x - 80
-2 = 13x - 80
78 = 13x
6 = x
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Know your log rules!!!
log a + log b = log (a * b)
log a - log b = log (a / b)
log (base x) a = b
converts to
a = x^b
2007-11-30 03:24:12 · answer #1 · answered by Mathematica 7 · 5⤊ 3⤋
log2(3x-2)-log2(x-5)=4
log2[(3x-2)/(x-5)] =4
base^log = number
so
2^4 = (3x-2)/(x-5)
16 = (3x-2)/(x-5)
(3x-2)/(x-5) = 16
Cross - multiply
3x-2 = 16x - 80
16x-80 = 3x-2
16x-3x = 80-2
13x=78
x=78/13
x=6
I would check the answer for validity:
log2(3x-2) - log2(x-5) = 4 and x=6
if
log2(18-2) - log2(6-5) =4
if log2(16) -log2(1) =4
if log2(16) -0 =4
if log2(2^4) = 4
if 4=4 True
2007-11-30 03:29:53 · answer #2 · answered by Sciman 6 · 1⤊ 1⤋
Let log stand for log base 2.
log (3x - 2) - log (x - 5) = 4
log [ (3x - 2) / (x - 5) ] = 4
(3x - 2) / (x - 5) = 2^4
3x - 2 = 16(x - 5)
3x - 2 = 16x - 80
78 = 13x
x = 6
2007-11-30 07:38:18 · answer #3 · answered by Como 7 · 3⤊ 1⤋