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.
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.
Personally, I would prefer 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.
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?
Hello, Latsabb!
Take logs, the base doesn't matter . . . Let's use natural logs.
We have: .
. . . . .
n . . .
. . .
. . .
n . . . . . . . .
Divide by . . . Note that it is negative.
. . . . . . . . . . . . .
. . . . Therefore: .
Hello again, Latsabb!
You should NOT take logs of a sum or difference.
It usually leads nowhere.
How would this be worked out if there was a negative exponent of x?
Let's say: .
Multiply by
Take logs: .
Therefore: .