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3^x = 5^y what is x/y ? Plz help . I'm terrible with Logs

Mathematics
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Bah, logs with different bases . . . Gimme a sec to remember how to do this.
Ok, got it. Take \(\large log_3\) of both sides.
Then you can use the logarithm power rule.

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Other answers:

Do you mind working please? @CliffSedge . I'm a little rusty on the rules
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?
*oops, never mind that = "3x" in the last line. I thought I deleted that.
I'm sorta catching on
Should be \(\large x=log_3(5^y)\)
Do you see that \(\large log_3(3^x)=x\) ?
The logarithm function tells you what the exponent is on that base.
So, since a logarithm is an exponent, and if you have a power-to-a-power, you multiply exponents, \(\large log_3(5^y)=y log_3(5)\) . I think you can get the rest from here.
|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.
I don't think the change of base is necessary, but it is good to know how to do such a thing.
I found this way to do it|dw:1351900905598:dw|
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
wow that's tough to read :(
That is an equivalent way, sure. It uses the same log rules but in a different order.
Sorry I just put them both to the 1/y
took the y'th root? Oh that's clever ^^
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 straight-forward way is easier to remember and just as efficient.

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