View Full Version : Solving an equation containing ln(x)
geobeeker
Feb 25, 2008, 01:49 PM
I have f'(x) = x(2ln(x) - 3) / (ln(x) - 1)^2 (which matches my book's answer)
My book says setting this equal to zero gives:
2ln(x) - 3 = 0
I think this implies that x / (ln(x) - 1)^2 equals 1
or (ln(x) - 1)^2 = x
but I don't understand why!
I need to see the steps.
Thanks,
Mr_am
Feb 25, 2008, 05:40 PM
if f'(x) is set to zero.. it implies x=o or 2ln(x)-3 is zero. Like saying A.B = zero means A is zero or B is zero. In this case A is x and 2ln(x)-3 is B. In this case we have A.B/C we are concerned with terms that make it zero.. obviously A or B being zero.
ebaines
Feb 26, 2008, 07:54 AM
I have f'(x) = x(2ln(x) - 3) / (ln(x) - 1)^2 (which matches my book's answer)
I think this implies that x / (ln(x) - 1)^2 equals 1
or (ln(x) - 1)^2 = x
but I don't understand why!
Sorry, but if 2 ln(x) - 3 = 0, this does not imply that x / (ln(x) - 1)^2 equals 1. Why do you think it does?
zenmeyang
Mar 10, 2008, 08:13 PM
If you say A.B / C = 0
Implies A= 0 or B = 0 But C must not equal to zero.
f'(x) = x(2ln(x) - 3) / (ln(x) - 1)^2 =0
That means x = 0 or 2 ln(x) - 3 = 0.
When x = 0, the denominator (ln(x) - 1)^2 also equal to 0.
It forms a 0/0, it is undefined.
That means x can't be zero.
So the only value makes f'(x) zero is 2 ln(x) - 3 = 0.
jasondelee
Feb 10, 2012, 11:52 PM
if a/b=o it implies a=0 but this (a=0)doesn't mean b=1 or 1/b=1 .so b can be any number except 0.the value of b doesn't affect the equation.cause a=0*b=0 for any value of b.