Asked by Em
Does anyone know how to solve,
ln((3x-2)/(x-1))<0
ln((3x-2)/(x-1))<0
Answers
Answered by
bobpursley
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
(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
Answered by
Reiny
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.
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.
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