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

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

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.2Then you can use the logarithm power rule.

3psilon
 2 years ago
Best ResponseYou've already chosen the best response.0Do you mind working please? @CliffSedge . I'm a little rusty on the rules

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.2Oh, 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?

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

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.2Should be \(\large x=log_3(5^y)\)

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

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.2The logarithm function tells you what the exponent is on that base.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.2So, 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.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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.

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

3psilon
 2 years ago
Best ResponseYou've already chosen the best response.0I found this way to do itdw:1351900905598:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0If 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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0wow that's tough to read :(

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

3psilon
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry I just put them both to the 1/y

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0took the y'th root? Oh that's clever ^^

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.2Makes 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.
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