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Thread: Exponential proof

  1. #1
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    Exponential proof

    My question is to prove the following:
    $\displaystyle (a^{b1})^{b2}=a^{b1b2}.$

    Any help would eb appreciated.
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  2. #2
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    Quote Originally Posted by thedoctor818 View Post
    My question is to prove the following:
    $\displaystyle (a^{b1})^{b2}=a^{b1b2}.$

    Any help would eb appreciated.
    $\displaystyle (a^{b1})^{b2} = e^{b2\ln{a^{b1}}} = e^{b2(\ln{e^{b1\ln{a}}})}$

    But, $\displaystyle \ln{e^{b1\ln{a}}} = b1\ln{a} $

    so, $\displaystyle (a^{b1})^{b2} = e^{b2(\ln{e^{b1\ln{a}}})} = e^{b1b2(\ln{a})} = a^{b1b2}.$
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    $\displaystyle (a^{b1})^{b2} = e^{b2\ln{a^{b1}}} = e^{b2(\ln{e^{b1\ln{a}}})}$

    But, $\displaystyle \ln{e^{b1\ln{a}}}) = b2b1\ln{a} $

    so, $\displaystyle (a^{b1})^{b2} = e^{b2(\ln{e^{b1\ln{a}}})} = e^{b1b2(\ln{a})} = a^{b1b2}.$
    Surely if the poster is asking the question they are at a point in their studies where they aren't supposed to use the natural logarithm. But, I guess without stating how or why he wants this proved your method is as good as any.
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  4. #4
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    What definitions do you have to work with? Certainly, for $\displaystyle b_1$ and $\displaystyle b_2$ positive integers, you can think of "$\displaystyle a^b_1$" as meaning "a multiplied by itself $\displaystyle b_1$ times" so that $\displaystyle (a^{b_1})^b_1$ is "(a multiplied by itself $\displaystyle b_1$ times) multiplied by itself $\displaystyle b_2$ times" so that $\displaystyle (a^{b_1})^{b_2}$ is a multiplied by itself a total of $\displaystyle b_1b_2$ times. Then, the standard procedure is to define define $\displaystyle a^b$ for b negative, a fraction, and an irrational number so that $\displaystyle (a^b_1)(a^b_2)= a^{b_1+ b_2}$, $\displaystyle (a^b_1)^b_2= a^{b_1b_2}$, and so that $\displaystyle a^b$ is a continuous function of b.
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  5. #5
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    Can I say the following:


    $\displaystyle (a^{b_1})^{b_2}=a^{b_1\;b_2}.$
    $\displaystyle =\exp[b_2\log(a^{b_1})].$
    $\displaystyle =\exp[b_2b_1\log(a)].$
    $\displaystyle =a^{b_2\;b_1}.$
    $\displaystyle =a^{b_1\;b_2}.$

    -Michael
    Last edited by Plato; Apr 23rd 2010 at 10:28 AM.
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  6. #6
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    Can I say the following:

    $\displaystyle
    (a^{b_1})^{b_2}=a^{b_1\;b_2}.\\
    =\exp[b_2\log(a^{b_1})].\\
    =\exp[b_2\;b_1\log(a)].\\
    =a^{b_2\;b_1}.\\
    =a^{b_1\;b_2}.
    $

    -Michael
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