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CliffSedgeBest ResponseYou've already chosen the best response.2
Bah, logs with different bases . . . Gimme a sec to remember how to do this.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Ok, got it. Take \(\large log_3\) of both sides.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Then you can use the logarithm power rule.
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
Do you mind working please? @CliffSedge . I'm a little rusty on the rules
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Oh, I suppose so. It's pretty easy.. Given \(\large 3^x = 5^y\) \(\large log_3\) both sides \(\large log_3(3^x) = log_3(5^y)\) \(\large x = 3^x = log_3(5^y)\) Understand so far?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
*oops, never mind that = "3x" in the last line. I thought I deleted that.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Should be \(\large x=log_3(5^y)\)
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Do you see that \(\large log_3(3^x)=x\) ?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
The logarithm function tells you what the exponent is on that base.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
So, since a logarithm is an exponent, and if you have a powertoapower, you multiply exponents, \(\large log_3(5^y)=y log_3(5)\) . I think you can get the rest from here.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
dw:1351900584391:dw Still having some trouble pimp or you figured this one out already. Figured I'd throw my 2 cents just in case :D Sorry that it's a little bit messy, hopefully you can follow it. The asterisks are log rules.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
I don't think the change of base is necessary, but it is good to know how to do such a thing.
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
I found this way to do itdw:1351900905598:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
If we wants to be able to plug it into a calculator it is necessary :) Depends what type of answer his teacher wants I suppose! :D
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
wow that's tough to read :(
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
That is an equivalent way, sure. It uses the same log rules but in a different order.
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
Sorry I just put them both to the 1/y
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
took the y'th root? Oh that's clever ^^
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Makes sense since it's looking for a ratio of exponents, but also takes a leap of imagination that might be hard to see. The straightforward way is easier to remember and just as efficient.
 one year ago
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