Common Logs: Inequalities

[Deo3560]

Ok, here is my Problem, I think I may have the answer, but I could use someone to check my work,

(1/4)^(3x)<6^(x-2)
(3x)ln(1/4)<(x-2)ln(6)
(3x)ln(1/4)<(x)ln(6)-(2)ln(6)
(3x)ln(1/4)-(x)ln(6)<(-2)ln(6)
x((3)ln(1/4)-ln(6))<(-2)ln(6)
Don't feel like typing this next step: Divide both sides by ((3)ln(1/4)-ln(6))
so then i got x?.6022

If this is wrong, can someone show me where I went wrong?

 

[tkhunny]

1) You ended up with an equality, rather than an inequality. What's up with that?
2) Did you consider the sign of your divisor before atttmpting division within an inequality?
3) Are you SURE you can use a logarithm in this way and not mess with the inequality? Does it ALWAYS work?
4) Why care ln(1/4) when the much lovelier -ln(4) could be used. This would also clarify the sign of -3ln(4)-ln(6).

Other than that, I think you have it! Good work.

 

[Deleted member 4993]

Ok, here is my Problem, I think I may have the answer, but I could use someone to check my work,

(1/4)^(3x)<6^(x-2)
(3x)ln(1/4)<(x-2)ln(6)
(3x)ln(1/4)<(x)ln(6)-(2)ln(6)
(3x)ln(1/4)-(x)ln(6)<(-2)ln(6)
x((3)ln(1/4)-ln(6))<(-2)ln(6)
Don't feel like typing this next step: Divide both sides by ((3)ln(1/4)-ln(6))
so then i got x?.6022

If this is wrong, can someone show me where I went wrong?

You have an "inequality" problem with "<" sign.

Your answer should be an inequality.

I don't feel like going through all the calculations either....

 

[Deo3560]

Thanks I suppose, I'll just test my luck after I make it an inequality, not sure if there is a sign for an Approx. Inequality.

 

[tkhunny]

The astute observer could change that denominator to this: 7ln(2) + ln(3)

It's a little simpler, I guess.

Why do you doubt?

 

[soroban]

Hello, Deo3560!

You did fine . . . but your answer needs clarification.

. . . . . . . . . . . . . .\(\displaystyle \left(\tfrac{1}{4}\right)^{3x} \;<\;6^{x-2}\)

. . . . . . . . . . \(\displaystyle 3x\ln\left(\tfrac{1}{4}\right) \;<\;(x-2)\ln(6)\)

. . . . . . . . . .\(\displaystyle 3x\ln\left(\tfrac{1}{4}\right) \;<\;x\ln(6)-2\ln(6)\)

. \(\displaystyle 3x\ln\left(\tfrac{1}{4}\right) - x\ln(6) \;<\; -2\ln(6)\)

\(\displaystyle x\bigg[3\ln\left(\tfrac{1}{4}\right)-\ln(6)\bigg] \;<\; -2\ln(6)\)

. . . . . . . . . . . . . . . .\(\displaystyle x \;<\;\frac{-2\ln(6)}{3\ln(\frac{1}{4}) - \ln(6)}\) . **


\(\displaystyle \text{So then I got: }\: x\:=\:0.6022\) .**

**
\(\displaystyle \text{Note that: }\:3\ln(\tfrac{1}{4}) - \ln(6) \:\;=\;\:\ln(\tfrac{1}{4})^3 - \ln(6) \:\;=\;\:\ln(\tfrac{1}{64}) - \ln(6) \:\;=\;\:\ln(\tfrac{1}{64}\cdot\tfrac{1}{6}) \:\;=\;\:\ln(\tfrac{1}{384})\;\text{ is }negative.\)

**
Is that "less than" or "greater than"?


I did it like this . . .

. . . . . . . . . . \(\displaystyle \left(\tfrac{1}{4}\right)^{3x} \;<\;6^{x-2}\)

. . . . . . . .\(\displaystyle 3x\ln(\tfrac{1}{4}) \;<\;(x-2)\ln(6)\)

. . . . . . \(\displaystyle 3x\ln(4^{-1}) \;<\;(x-2)\ln(6)\)

. . . . . . .\(\displaystyle -3x\ln(4) \;<\;(x-2)\ln(6)\)

. . . . . . .\(\displaystyle -3x\ln(4) \;<\;x\ln(6) - 2\ln(6)\)

.\(\displaystyle -3x\ln(4) - x\ln(6) \;<\:-2\ln(6)\)

\(\displaystyle -x\bigg[3\ln(4) + \ln(6)\bigg] \;<\;-2\ln(6)\)

. . . . . . . . . . . \(\displaystyle -x \;<\;\frac{-2\ln(6)}{3\ln(4) + \ln(6)}\)


\(\displaystyle \text{Multiply by -}1: \;x \;>\;\frac{2\ln(6)}{3\ln(4) + \ln(6)} \;=\; 0.602207057\)


\(\displaystyle \text{Therefore: }\;x \;>\;0.602\)