Logarithm inequality with different constants?

The problem is as such:

5*3^{x}>12*5^{x}

I have been mulling over this thing for quite some time now. Normally I would start by dividing both sides by the 5, but that doesnt isolate the x, so that doesnt seem to be the way to go. So I turned everything into logs.

log 5 + x log 3 > log 12 + x log 5

From here, I have tried just about everything I could come up with. Subtract around so that you get the x logs on one side, and the non-variable logs on the other side, but then they dont seem to match up. Can someone let me know where I go from here? I am sketchy on what is allowed with this. Can I simply have x log 3 - x log 5, and combine them somehow, even though one is a log 3, and the other is a log 5? I feel like my head is swimming from all the conversions back and forth between log and not log formats.

Re: Logarithm inequality with different constants?

Quote:

Originally Posted by

**Latsabb** log 5 + x log 3 > log 12 + x log 5

In this equation log is to the same (arbitrary) base. This is a linear equation in x.

Re: Logarithm inequality with different constants?

from $\displaystyle log10(5)+xlog10(3)>log10(12)+xlog10(5)$

$\displaystyle xlog10(3)-xlog10(5)>log10(12)-log10(5)$

$\displaystyle x(log10(3)-log10(5))>log10(12)-log10(5)$

$\displaystyle x<(log10(12)-log10(5))/(-log10(5)-log(3))$

$\displaystyle x<-(log10(12/5)/(log10(5/3))$

Re: Logarithm inequality with different constants?

Ok, so then I break that down to

x log 3/5 > log 12/5

So I could then isolate x by dividing by log 3/5 on both sides, and flipping the inequality sign. However, the book states that the answer must be in "exact" form. (ie log 2/3, not 0.63, for example)

Just to make sure, I can just call that (log 12 - log 5) / (log 3 - log 5) correct? That seems like it should be ok in my head, but like I said, I have been mulling over things for a while, and I am kind of burnt.

Re: Logarithm inequality with different constants?

Personally, I would prefer $\displaystyle x> \frac{log(12/5)}{log(3/5}$ rather than using the differences. Note the ">" rather than "<". Since 3/5< 1, log(3/5) is negative and dividing both sides by a negative number reverses the direction of the inequality.

Re: Logarithm inequality with different constants?

But since the problem originated with the >, shouldnt it be flipped to <?

Also, how would this be worked out if there was a negative exponent of x? Lets say:

3^{x} - 4 * 3^{-x} > 0

It seems as if I put those into logs, it would turn into

x log 3 + (-x)log 3 > log 4

Would that be the correct way of doing it? Because that looks like the two x log 3 groups should cancel each other out. Even if I use x log 1/3, wouldnt that then be x log (3)(1/3) and just become log 1, which would be log 0?

Re: Logarithm inequality with different constants?

Hello, Latsabb!

Quote:

$\displaystyle \text{Solve for }x\!:\;\;5\cdot3^x\:>\:12\cdot5^x$

Take logs, the base doesn't matter . . . Let's use natural logs.

We have: .$\displaystyle \ln(5\cdot3^x) \;>\;\ln(12\cdot5^x) $

. . . . . $\displaystyle \ln(5) + \ln(3^x) \;>\;\ln(12) + \ln(5^x)$

n . . . $\displaystyle \ln(5) + x\ln(3) \;>\;\ln(12) + x\ln(5)$

. . . $\displaystyle x\ln(3) - x\ln(5) \;>\;\ln(12) - \ln(5)$

. . . $\displaystyle x\big[\ln(3) - \ln(5)\big] \;>\;\ln(12) - \ln(5)$

n . . . . . . . . $\displaystyle x\ln(\tfrac{3}{5}) \;>\;\ln(\tfrac{12}{5})$

Divide by $\displaystyle \ln(\tfrac{3}{5})$ . . . Note that it is *negative*.

. . . . . . . . . . . . . $\displaystyle x \;{\color{red}<} \;\frac{\ln(\frac{12}{5})}{\ln(\frac{3}{5})} $

. . . . Therefore: .$\displaystyle x \;<\;\text{-}1.713830898$

Re: Logarithm inequality with different constants?

Hello again, Latsabb!

You should NOT take logs of a sum or difference.

It usually leads nowhere.

Quote:

How would this be worked out if there was a negative exponent of x?

Let's say: .$\displaystyle 3^x - 4\cdot3^{-x} \;>\; 0$

Multiply by $\displaystyle 3^x\!:\;\;3^{2x} - 4 \;>\;0 \quad\Rightarrow\quad 3^{2x} \;>\;4$

Take logs: .$\displaystyle \ln(3^{2x}) \;>\;\ln(4) \quad\Rightarrow\quad 2x\ln(3) \;>\;\ln(4)$

Therefore: .$\displaystyle x \;>\;\frac{\ln(4)}{2\ln(3)} \;\approx\;0.630929754$