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Thread: A simple algebra exercise

  1. #1
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    A simple algebra exercise

    $\displaystyle 3^{x} = 2$

    $\displaystyle 27^{-x} = ?$


    The result is

    $\displaystyle \dfrac{1}{8}$

    How do I get to the answer?
    Thanks!
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  2. #2
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    Apply these 2 rules:

    If a^p = b then p = log(b) / log(a)

    a^(-p) = 1 / a^p
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  3. #3
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    Quote Originally Posted by klik11 View Post
    $\displaystyle 3^{x} = 2$

    $\displaystyle 27^{-x} = ?$


    The result is

    $\displaystyle \dfrac{1}{8}$

    How do I get to the answer?
    Thanks!
    Slightly different way that avoids logs.

    $\displaystyle 27^{-x}=(3^3)^{-x}=3^{-3x}=(3^x)^{-3}=2^{-3}=\frac{1}{8}$
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  4. #4
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    Thanks!

    I have one more exercise like this though, and I still can't solve it.

    $\displaystyle 3^{2x+8} = b^{8}$

    $\displaystyle 9^{-x} = ?$


    the result is

    $\displaystyle (\frac{b}{3})^{8}$
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  5. #5
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    Quote Originally Posted by klik11 View Post
    Thanks!

    I have one more exercise like this though, and I still can't solve it.

    $\displaystyle 3^{2x+8} = b^{8}$

    $\displaystyle 9^{-x} = ?$


    the result is

    $\displaystyle (\frac{b}{3})^{8}$
    $\displaystyle 3^{2x+8}=3^{2x}\cdot3^8=9^x\cdot3^8$

    continue..

    Actually I get $\displaystyle (\frac{3}{b})^{8}$
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  6. #6
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    Yes my mistake, it's

    $\displaystyle (\frac{3}{b})^{8}$

    Actually, I already figured that 9^x equals e^2x. I have no idea how to continue the exercise...
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  7. #7
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    Quote Originally Posted by klik11 View Post
    Yes my mistake, it's

    $\displaystyle (\frac{3}{b})^{8}$

    Actually, I already figured that 9^x equals e^2x. I have no idea how to continue the exercise...
    e^2x?

    Continuing from where I left off above

    $\displaystyle 9^x\cdot3^8=b^8$

    $\displaystyle 9^x=(\frac{b}{3})^8$

    $\displaystyle 9^{-x}=(\frac{3}{b})^8$
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  8. #8
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    3^2x *

    Thanks!!!
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