Here's the question you clicked on:
cathyangs
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 (7lnx-7ln2)/6ln2
Hint: \[ (2x)^n = 2^n x^n. \]Would you mind telling us what, exactly, you are having trouble with?
It's not necessary to do that, cathy. It's better the way the problem statement is given. Exponents are what we need.
What I just wrote is what the problem asks, I just wanted to clear any confusions.
Oh. I thought you were ... okay. The way you ahve it written is how you want it.
Maybe you should show me your work? Sorry for the double-post. have*
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?
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.
Nope. x has to be known. I think you're good! :)
is my answer of (7lnx-7ln2)/6ln2 alright, then?
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.
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?
Hm.... I got lnx=(n6/7 +1) ln2 then (n6/7+`)= lnx/ln2 then n=(lnx/ln2 -1)7/6 which =7(lnx-ln2)/6ln2 =7lnx-7ln2/6ln2
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
Thanks for the medal, anyways. Really appreciate the reward for my effort! :D