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heathernelly
 2 years ago
;)))
heathernelly
 2 years ago
;)))

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jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1Similar Problem: \[\Large \log_{3}(60)  \log_{3}(12)\] \[\Large \log_{3}\left(\frac{60}{12}\right)\] \[\Large \log_{3}(5)\] So \[\Large \log_{3}(60)  \log_{3}(12) = \log_{3}(5)\]

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0that's all i have to do?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1pretty much, but just use your numbers in place of mine

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, thanks! do you know how to do the other one?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1in general \[\Large \log_{b}(x)  \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)\]

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1what do 16 and 64 have in common

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0they both have a 6 in them, both even #'s

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1good, but in terms of factors

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.02^6? i don't know :S :/

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1both are powers of 2 or powers of 4

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1so 16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1well you need to rewrite both sides to have the same base

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1for 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)

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1so you're using this idea to solve the second problem

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0ohhh, well that's pretty simple! thank youu

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0actually waait, it's not! haha.. i just got confused

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1where are you stuck

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

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1well if b^x = b^y, then x = y

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0i'm confused because you did 2 different problems..:/

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1that's just an example

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0okay, 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''

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1oh i meant to say 16 = 4^2, my bad

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.11/16=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

heathernelly
 2 years ago
Best ResponseYou've already chosen the best response.0ahh, got it :) thanks a lot!!!
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