At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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.
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.