ln (3x-2/x-1) <0
Can I re-write this as (3x-2/x-1)<e^0
And e^0 is 1?
And then I can solve for x?
I will assume you meant:
ln ( (3x-2)/(x-1) ) < 0
I would establish the boundary by letting
ln ( (3x-2)/(x-1) ) = 0
then (3x-2)/(x-1) = e^0 , yes e^0 = 1
(3x-2)/(x-1) = 1
3x-2 = x-1
2x = 1
x = 1/2
So we need either x < 1/2 or x > 1/2
But remember x = 1 is not allowed or else we are dividing by zero
Furthermore, remember we can only take logs of positive numbers
so (3x-2)/(x-1) > 0 , clearly x ≠ 1
and a quick analysis will show that x > 2/3 for the expression >0 before we even take the log
at x=2/3, we would be taking ln(0) which is undefined
so
1/2 ≤ x ≤ 2/3
Wolfram's graph shows the grap below the x - axis in blue and
confirms my solution
http://www.wolframalpha.com/input/?i=plot+y+%3D+ln+%28+%283x-2%29%2F%28x-1%29+%29+
GREAT QUESTION!