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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.

Do you mind working please? @CliffSedge . I'm a little rusty on the rules

*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.

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|

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 ^^