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Similar Problem: \[\Large \log_{3}(60) - \log_{3}(12)\] \[\Large \log_{3}\left(\frac{60}{12}\right)\] \[\Large \log_{3}(5)\] So \[\Large \log_{3}(60) - \log_{3}(12) = \log_{3}(5)\]
that's all i have to do?
pretty much, but just use your numbers in place of mine

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Other answers:

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
for instance, say you had 2^x = 16 you would rewrite 16 as 2^4 to get 2^x = 2^4 and you can see that x must be 4 (since both bases are 2)
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
okay, that makes sense .. but i'm still confused on what to do after what you said ''16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3''
oh i meant to say 16 = 4^2, my bad
1/16=64^(4x-3) 1/(4^2) = (4^3)^(4x-3) ... since 16 = 4^2 and 64 = 4^3 4^(-2) = 4^(3*(4x-3)) so because the bases are now the same, we can say that the exponents must be the same so... -2 = 3*(4x-3) solve for x
ahh, got it :) thanks a lot!!!
np

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