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jim_thompson5910Best ResponseYou've already chosen the best response.1
Similar Problem: \[\Large \log_{3}(60)  \log_{3}(12)\] \[\Large \log_{3}\left(\frac{60}{12}\right)\] \[\Large \log_{3}(5)\] So \[\Large \log_{3}(60)  \log_{3}(12) = \log_{3}(5)\]
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
that's all i have to do?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
pretty much, but just use your numbers in place of mine
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
yeah, thanks! do you know how to do the other one?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
in general \[\Large \log_{b}(x)  \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)\]
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
what do 16 and 64 have in common
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
they both have a 6 in them, both even #'s
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
good, but in terms of factors
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
2^6? i don't know :S :/
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
both are powers of 2 or powers of 4
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
so 16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
well you need to rewrite both sides to have the same base
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
so you're using this idea to solve the second problem
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
ohhh, well that's pretty simple! thank youu
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
actually waait, it's not! haha.. i just got confused
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
where are you stuck
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
and you can see that x must be 4 (since both bases are 2) ??
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
well if b^x = b^y, then x = y
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
i'm confused because you did 2 different problems..:/
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
that's just an example
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
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''
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
oh i meant to say 16 = 4^2, my bad
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
1/16=64^(4x3) 1/(4^2) = (4^3)^(4x3) ... since 16 = 4^2 and 64 = 4^3 4^(2) = 4^(3*(4x3)) so because the bases are now the same, we can say that the exponents must be the same so... 2 = 3*(4x3) solve for x
 one year ago

heathernellyBest ResponseYou've already chosen the best response.0
ahh, got it :) thanks a lot!!!
 one year ago
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