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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

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  1. Ahaanomegas
    • 3 years ago
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    Hint: \[ (2x)^n = 2^n x^n. \]Would you mind telling us what, exactly, you are having trouble with?

  2. Ahaanomegas
    • 3 years ago
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    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
    • 3 years ago
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    What I just wrote is what the problem asks, I just wanted to clear any confusions.

  4. Ahaanomegas
    • 3 years ago
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    Oh. I thought you were ... okay. The way you ahve it written is how you want it.

  5. Ahaanomegas
    • 3 years ago
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    Maybe you should show me your work? Sorry for the double-post. have*

  6. Ahaanomegas
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Nope. x has to be known. I think you're good! :)

  9. cathyangs
    • 3 years ago
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    is my answer of (7lnx-7ln2)/6ln2 alright, then?

  10. Ahaanomegas
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    XD It's fine.

  15. Ahaanomegas
    • 3 years ago
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    Thanks for the medal, anyways. Really appreciate the reward for my effort! :D

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