A community for students.
Here's the question you clicked on:
 0 viewing
heathernelly
 3 years ago
;)))
heathernelly
 3 years ago
;)))

This Question is Closed

jim_thompson5910
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0that's all i have to do?

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

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

jim_thompson5910
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1what do 16 and 64 have in common

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

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

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

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

jim_thompson5910
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1well you need to rewrite both sides to have the same base

jim_thompson5910
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1so you're using this idea to solve the second problem

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

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

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

heathernelly
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1well if b^x = b^y, then x = y

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

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

heathernelly
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1oh i meant to say 16 = 4^2, my bad

jim_thompson5910
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0ahh, got it :) thanks a lot!!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.