# Solving an inequality.

• Apr 6th 2011, 03:12 AM
jessmari86
Solving an inequality.
problem is solve the inequality:

2^(2x-1)<3^(3x-2)
2 to the 2x-1 power<3 to the 3x-2 power

1) took ln of 2 and 3

(2x-1)ln2<(3x-2)ln3

2) factored

2x*ln(2)-ln(2)<3x*ln(3)-2ln(3)

3) added ln2 to both sides

2x*ln(2)<3x*ln(3)+ln(2)-2ln(3)

I am stuck as to what to do next i know i must isolate the x but dont see how it will work and still be able to come up with the answer provided by a online solver.

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

please any help would be great!!!!! please if you could explain thank you(Thinking)
• Apr 6th 2011, 03:33 AM
Quote:

Originally Posted by jessmari86
problem is solve the inequality:

2^(2x-1)<3^(3x-2)
2 to the 2x-1 power<3 to the 3x-2 power

1) took ln of 2 and 3

(2x-1)ln2<(3x-2)ln3

2) factored

2x*ln(2)-ln(2)<3x*ln(3)-2ln(3)

3) added ln2 to both sides

2x*ln(2)<3x*ln(3)+ln(2)-2ln(3)

I am stuck as to what to do next i know i must isolate the x but dont see how it will work and still be able to come up with the answer provided by a online solver.

To isolate x, you need x on one side.
Subtract 3xln3 from both sides, then factor out x next.
Finally divide both sides by the single remaining multiplier of x

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

please any help would be great!!!!! please if you could explain thank you(Thinking)

.
• Apr 6th 2011, 03:55 AM
pickslides
$2x\ln 2-\ln 2< 3x \ln 2 -2\ln 2$

$2x\ln 2< 3x \ln 2 -2\ln 2 +\ln 2$

$2x\ln 2- 3x \ln 2< -2\ln 2 +\ln 2$

$x(2\ln 2- 3 \ln 2) < -2\ln 2 +\ln 2$

$\displaystyle x < \frac{-2\ln 2 +\ln 2}{2\ln 2- 3 \ln 2}$

$\displaystyle x < \frac{\ln 2-2\ln 2 }{2\ln 2- 3 \ln 2}$
• Apr 6th 2011, 04:36 AM
amul28
pickslides, the RHS in the question has

$3^{3x-1}$

So when you take $ln$ it will be
$(3x-1)ln3$

$