Does anyone know how to solve,

ln((3x-2)/(x-1))<0

2 answers

take the antilog of each side
(3x-2)/(x-1)<1

3x-2 < x-1
2x<1
x<.5

check:
ln(1.5-2)/(.5-1)
ln((-.5/-.5)=ln(1)=0
ln((3x-2)/(x-1))<0
using basic definition of logs ....
ln(3x-2) - ln(x-1) < 0

by definition:
x > 2/3 and x> 1
so x > 1 , before we even proceed further.

but bob found that if we leave it as
ln((3x-2)/(x-1)) < 0
we do get a solution of x < 1/2

thus x < 1/2 AND x > 1

no solution.