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Thread: e, ln(x)

  1. #1
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    e, ln(x)

    My question has to do with simplifying.

    $\displaystyle e^(-9ln(x)) = 1/e^(9ln(x)) $

    What does this simplify to?

    [Math] 1/9x [/tex] ?
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  2. #2
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    Quote Originally Posted by KyleC View Post
    My question has to do with simplifying.

    $\displaystyle e^{(-9ln(x))} = 1/e^{(9ln(x))} $

    What does this simplify to?

    [Math] 1/9x [/tex] ?

    $\displaystyle kln(a) = ln(a^k)$

    $\displaystyle e^{-9ln(x)} = e^{ln(x^{-9})}$

    $\displaystyle \frac{1}{e^{9ln(x)}} = \frac{1}{e^{ln({x^9})} = $

    Combining and cancelling:

    $\displaystyle x^{-9} = \frac{1}{x^9}$

    $\displaystyle x^9$ to give $\displaystyle 1=1$
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  3. #3
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    Ahh, Thank you very much.
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  4. #4
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    Quote Originally Posted by KyleC View Post
    $\displaystyle e^{-9ln(x)} = 1/e^{9ln(x)} $

    What does this simplify to?
    A basic property of logs leads to the conclusion that, for any base "b", the following holds:

    . . . . .$\displaystyle b^{\log_b(m)}\, =\, m$

    Applying a log rule inside that power to put that -9 back inside the log, we get a power of:

    . . . . .$\displaystyle -9\ln(x)\, =\, \ln(x^{-9})\, =\, \ln\left(\frac{1}{x^9}\right)$

    ...by the rule for negative exponents.

    Now apply the first rule above to simply the exponential expression.
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