Does anyone know how to solve,
ln((3x-2)/(x-1))<0
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.
To solve the inequality ln((3x-2)/(x-1)) < 0, follow these steps:
Step 1: Set up the inequality without the logarithm.
(3x-2)/(x-1) < 1
Step 2: Find the critical points by setting the expression equal to zero and solving for x.
(3x-2)/(x-1) = 0
To solve this equation, set the numerator equal to zero:
3x-2 = 0
Solve for x:
3x = 2
x = 2/3
Now set the denominator equal to zero:
x-1 = 0
Solve for x:
x = 1
Therefore, the critical points are x = 2/3 and x = 1.
Step 3: Construct a number line and test intervals.
On the number line, place the critical points:
- - - - - - - - - -
1 2/3
Choose a test point in each interval to determine the sign of the expression.
For the interval (-∞, 2/3), choose x = 0
(3(0)-2)/(0-1) = 2/1 = 2
Since 2 > 1, the expression is positive in this interval.
Therefore, this part of the number line is not a solution.
For the interval (2/3, 1), choose x = 3/4
(3(3/4)-2)/(3/4-1) = (9/4 - 2)/(-1/4) = (9/4 - 8/4)/(-1/4) = (1/4)/(-1/4) = -1
Since -1 < 1, the expression is negative in this interval.
Therefore, this part of the number line is a valid solution.
For the interval (1, ∞), choose x = 2
(3(2)-2)/(2-1) = (6-2)/1 = 4
Since 4 > 1, the expression is positive in this interval.
Therefore, this part of the number line is not a solution.
Step 4: Express the solution.
Based on the number line analysis, the solution to ln((3x-2)/(x-1)) < 0 is:
2/3 < x < 1