## cathyangs 3 years ago 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

1. Ahaanomegas

Hint: $(2x)^n = 2^n x^n.$Would you mind telling us what, exactly, you are having trouble with?

2. Ahaanomegas

It's not necessary to do that, cathy. It's better the way the problem statement is given. Exponents are what we need.

3. cathyangs

What I just wrote is what the problem asks, I just wanted to clear any confusions.

4. Ahaanomegas

Oh. I thought you were ... okay. The way you ahve it written is how you want it.

5. Ahaanomegas

Maybe you should show me your work? Sorry for the double-post. have*

6. Ahaanomegas

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?

7. cathyangs

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.

8. Ahaanomegas

Nope. x has to be known. I think you're good! :)

9. cathyangs

is my answer of (7lnx-7ln2)/6ln2 alright, then?

10. Ahaanomegas

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.

11. Ahaanomegas

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?

12. cathyangs

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

13. Ahaanomegas

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

14. cathyangs

XD It's fine.

15. Ahaanomegas

Thanks for the medal, anyways. Really appreciate the reward for my effort! :D

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