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that's all i have to do?

pretty much, but just use your numbers in place of mine

yeah, thanks! do you know how to do the other one?

in general
\[\Large \log_{b}(x) - \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)\]

what do 16 and 64 have in common

they both have a 6 in them, both even #'s

good, but in terms of factors

2^6? i don't know :S :/

both are powers of 2 or powers of 4

ahh, yeah

so
16 = 2^4
64 = 2^6
and
16 = 4^1
64 = 4^3

that's it?

well you need to rewrite both sides to have the same base

so you're using this idea to solve the second problem

ohhh, well that's pretty simple! thank youu

actually waait, it's not! haha.. i just got confused

with what

where are you stuck

and you can see that x must be 4 (since both bases are 2) ??

well if b^x = b^y, then x = y

i'm confused because you did 2 different problems..:/

that's just an example

oh i meant to say 16 = 4^2, my bad

ahh, got it :) thanks a lot!!!