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cathyangs
Group Title
For what value of n does
((2x)^(1/2)*(2x^2)^(1/3))/2^2 = 2^n
How do I find n? The best I've come up with is (7lnx7ln2)/6ln2
 2 years ago
 2 years ago
cathyangs Group Title
For what value of n does ((2x)^(1/2)*(2x^2)^(1/3))/2^2 = 2^n How do I find n? The best I've come up with is (7lnx7ln2)/6ln2
 2 years ago
 2 years ago

This Question is Closed

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Hint: \[ (2x)^n = 2^n x^n. \]Would you mind telling us what, exactly, you are having trouble with?
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
It's not necessary to do that, cathy. It's better the way the problem statement is given. Exponents are what we need.
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.0
What I just wrote is what the problem asks, I just wanted to clear any confusions.
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Oh. I thought you were ... okay. The way you ahve it written is how you want it.
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Maybe you should show me your work? Sorry for the doublepost. have*
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
I have: \[ \dfrac {(2x)^{1/2} (2x)^{1/3}}{2^2} = 2^n \rightarrow \dfrac {(2x)^{5/6}}{2^2} = 2^n \rightarrow \dfrac {x^{5/6}}{2^{7/6}} = 2^n. \]So: \[ x^{5/6} = 2^{n + \tfrac {7}{6}}. \]Do you see where this is heading?
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.0
First I multiplied by 2^2,so the right side is 2^(n+2) I expanded the exponents, so it because 2^1/2*x^1/2*2^1/3*x^2/3=2^(n+2) Thankts what I did. But the only solution I can find is in terms of x. I want to know if there's a way to find the value of n not in terms of x.
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Nope. x has to be known. I think you're good! :)
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.0
is my answer of (7lnx7ln2)/6ln2 alright, then?
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
I don't think so but I may be mistaken. From my work above, the answer is: \[ n = \boxed {\dfrac {5 \ln x}{6 \ln 2}  \dfrac {7}{6}} \]This may be a different form of your answer. Let me think. I'm posting this for now so that you get some time to think, as well.
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Oh. I see. You have the same answer as I do, except that you have a 7 where I have a 5. Do you see your mistake?
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.0
Hm.... I got lnx=(n6/7 +1) ln2 then (n6/7+`)= lnx/ln2 then n=(lnx/ln2 1)7/6 which =7(lnxln2)/6ln2 =7lnx7ln2/6ln2
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
I apologize. I just realized that you were correct! I had a 5/6 for the exponent on the top and a 12/6 on the bottom. I forgot to make that difference 7/6. Good job and sorry about that! There goes my chance of getting a medal for this thread. =P
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.0
XD It's fine.
 2 years ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Thanks for the medal, anyways. Really appreciate the reward for my effort! :D
 2 years ago
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