Solve logarithmic equation: log(x–5) – log(3–x) = log1
2007-04-23 09:56:45 · 9 answers · asked by Michael S 1 in Science & Mathematics ➔ Mathematics
use the law of logs
log [(x-5)/(3-x)] = log1
(x - 5) / (3 - x) = 1
x - 5 = 3 - x
x = 4
2007-04-23 09:59:56 · answer #1 · answered by Dr D 7 · 1⤊ 1⤋
log(x–5) – log(3–x) = log1(x-5/(3-x=log 1
x>5 and x<3
(x-5/(3-x=1
or x-5=3-x or 2x=-8 or x=4 . so no solution
2007-04-23 10:07:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
log(x–5) – log(3–x) = log1
x-5> 0 so x >5
3–x>0 so x<3
So no solution.
You might think
log(x–5) – log(3–x) = log1
log(x–5) – log(3–x) = 0
log(x–5) = log(3–x)
x-5 = 3-x so x = 4 Nevertheless this is not a solution.
Th
2007-04-23 10:03:06 · answer #3 · answered by Thermo 6 · 0⤊ 0⤋
log(x-5) - log (3-x) = log 1
Use the properties of log's:
log { (x-5) / (3-x) } = log 1
raise each side 10^... to eliminate the logs
(x-5)/(3-x) = 1, solve like a normal equation:
x-5 = 3 - x
2007-04-23 10:01:13 · answer #4 · answered by Mαtt 6 · 1⤊ 0⤋
We know that log(a)-log(b)=log(a/b)
Applying the same here we get
log((x-5)/(3-x))log 1
Taking antilog on both sides we get
x-5/3-x=1 or x-5=3-x
or x=4
2007-04-23 10:02:06 · answer #5 · answered by Napster 2 · 0⤊ 0⤋
log 1 = 0
log(x - 5) = log(3 - x)
x - 5 = 3 - x
2x = 8
x = 4
2007-04-23 10:04:59 · answer #6 · answered by Como 7 · 0⤊ 1⤋
log2 [(2x + a million)(x + 4)] = 2 2^( log2 [(2x + a million)(x + 4)] ) = 2^2 2x^2 + 8x + x + 4 = 4 2x^2 + 9x = 0 x(2x + 9) = 0 2x = - 9 x = - 9/2 x = 0 log2 [(-9 + a million)(-9/2 + 4)] = log2 [(-8)(-a million/2)] = log2 [4] = log2 [2^2] = 2log 2 [2] = 2
2016-10-13 07:25:52 · answer #7 · answered by ghil 4 · 0⤊ 0⤋
log(x-5/3-x)=log1
x-5/3-x=1
x-5=3-x
x=4
2007-04-23 10:00:06 · answer #8 · answered by Anonymous · 0⤊ 0⤋
log1=0
log[(x-5)/(3-x)]=0
so (x-5)/(3-x)=1
x-5=3-x
2x=8
x=4
2007-04-23 10:01:55 · answer #9 · answered by Anonymous · 0⤊ 0⤋