Isnt the answer x=2??
2007-06-03 08:35:48 · 6 answers · asked by ilovesoccer 1 in Science & Mathematics ➔ Mathematics
Yes, but when x=2, 5-3x=-1 and log -1 is not defined.
2007-06-03 08:39:14 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
As negative numbers don´t have log
5-3x>0 so x<5/3 and 4x-9>0 so x>9/4
With this last condition
5-3x=4x-9 and x= 2 which does not comply the condition and so is not solution
2007-06-03 15:43:45 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
log(-1) = log(exp(i(2k+1)pi)) (k=integers)
log(-1) = i(2k+1)pi
Each k is a separate branch. The primary solution is for k=0 and is log(-1) = ipi. The others are due to the 2pi periodicity.
There are no real solutions, just the imaginary solution at x=2 which is infinitely branched.
2007-06-03 15:55:44 · answer #3 · answered by jcsuperstar714 4 · 0⤊ 0⤋
In the field of real numbers your answer has no solution.
With complex numbers , â(-1) = i and log(-1) = iÏ. So yes, the solution would be x = 2.
2007-06-03 16:04:36 · answer #4 · answered by J w 2 · 0⤊ 0⤋
log (5-3x)=log (4x-9)
5-3x = 4x-9
5+9 = 4x+3x
14 = 7x
x = 14/7
x = 2
Sub x=2 into each side:
LHS = log -1
RHS = log -1 (NA)
So this equation has no solution.
2007-06-03 15:42:04 · answer #5 · answered by Kemmy 6 · 1⤊ 0⤋
http://www.coolmath.com
2007-06-03 15:38:57 · answer #6 · answered by Anonymous · 0⤊ 2⤋