Delete
Share
This Question is Closed
jim_thompson5910
Best Response
You'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 \[\Large \log_{3}(60) - \log_{3}(12) = \log_{3}(5)\]
heathernelly
Best Response
You've already chosen the best response.
0
that's all i have to do?
jim_thompson5910
Best Response
You've already chosen the best response.
1
pretty much, but just use your numbers in place of mine
heathernelly
Best Response
You've already chosen the best response.
0
yeah, thanks! do you know how to do the other one?
jim_thompson5910
Best Response
You've already chosen the best response.
1
in general
\[\Large \log_{b}(x) - \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)\]
jim_thompson5910
Best Response
You've already chosen the best response.
1
what do 16 and 64 have in common
heathernelly
Best Response
You've already chosen the best response.
0
they both have a 6 in them, both even #'s
jim_thompson5910
Best Response
You've already chosen the best response.
1
good, but in terms of factors
heathernelly
Best Response
You've already chosen the best response.
0
2^6? i don't know :S :/
jim_thompson5910
Best Response
You've already chosen the best response.
1
both are powers of 2 or powers of 4
heathernelly
Best Response
You've already chosen the best response.
0
ahh, yeah
jim_thompson5910
Best Response
You've already chosen the best response.
1
so
16 = 2^4
64 = 2^6
and
16 = 4^1
64 = 4^3
heathernelly
Best Response
You've already chosen the best response.
0
that's it?
jim_thompson5910
Best Response
You've already chosen the best response.
1
well you need to rewrite both sides to have the same base
jim_thompson5910
Best Response
You'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)
jim_thompson5910
Best Response
You've already chosen the best response.
1
so you're using this idea to solve the second problem
heathernelly
Best Response
You've already chosen the best response.
0
ohhh, well that's pretty simple! thank youu
heathernelly
Best Response
You've already chosen the best response.
0
actually waait, it's not! haha.. i just got confused
jim_thompson5910
Best Response
You've already chosen the best response.
1
with what
jim_thompson5910
Best Response
You've already chosen the best response.
1
where are you stuck
heathernelly
Best Response
You've already chosen the best response.
0
and you can see that x must be 4 (since both bases are 2) ??
jim_thompson5910
Best Response
You've already chosen the best response.
1
well if b^x = b^y, then x = y
heathernelly
Best Response
You've already chosen the best response.
0
i'm confused because you did 2 different problems..:/
jim_thompson5910
Best Response
You've already chosen the best response.
1
that's just an example
heathernelly
Best Response
You'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''
jim_thompson5910
Best Response
You've already chosen the best response.
1
oh i meant to say 16 = 4^2, my bad
jim_thompson5910
Best Response
You've already chosen the best response.
1
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
heathernelly
Best Response
You've already chosen the best response.
0
ahh, got it :) thanks a lot!!!