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Math Help - e, ln(x)

  1. #1
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    e, ln(x)

    My question has to do with simplifying.

     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.

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

    What does this simplify to?

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

    kln(a) = ln(a^k)

    e^{-9ln(x)} = e^{ln(x^{-9})}

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

    Combining and cancelling:

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

    x^9 to give 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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    Talking

    Quote Originally Posted by KyleC View Post
     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:

    . . . . . 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:

    . . . . . -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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