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Exponential equation help

  1. Jul 17, 2008 #1
    Hello all, I can't believe I having a hard time with this problem but...I am :cry:


    1. 2^(x^2) X 4^(2x) = 1/8



    2. make the bases the same?



    3.
    2^(x^2) X 2^((2x)^2) = 1/8
    2^(3x^(2)) = 1/8
    3x^(2)log2=log1/8
    (log1/8)/(log2) = 3x^(2)
    x^(2) = -1


    I know I am doing it wrong. Any help would be greatly appreciated. Thanks,
    Baba
     
  2. jcsd
  3. Jul 17, 2008 #2

    rock.freak667

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    [tex](2^{2x})^2=2^{4x}[/tex]
     
  4. Jul 17, 2008 #3

    dynamicsolo

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    The general rule for "exponentiated exponents" is

    [tex](a^b)^c = a^{(b \cdot c)[/tex], for a positive base a .

    For this problem, you will end up with a quadratic equation in x, with two solutions.
     
    Last edited: Jul 17, 2008
  5. Jul 17, 2008 #4
    great! thanks so much. Just a stupid mistake on my part, I knew that :uhh:
     
  6. Jul 18, 2008 #5
    I believe you end up with two imaginary solutions; My guess is that you should just show your work until you run into negative square root.
     
  7. Jul 18, 2008 #6

    rock.freak667

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    You actually get two real roots.
     
  8. Jul 18, 2008 #7
    I found the same, you should get two real roots.
     
  9. Jul 18, 2008 #8
    Oh, I see what I did wrong, I had [tex]2^{x^2}(2^x + 2^x)=1/8[/tex], instead of [tex] 2^{x^2}*2^x*2^x=1/8[/tex]
     
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