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Math Help - Exponential proof

  1. #1
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    Exponential proof

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

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

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

    so,  (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
     (a^{b1})^{b2} = e^{b2\ln{a^{b1}}} = e^{b2(\ln{e^{b1\ln{a}}})}

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

    so,  (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 b_1 and b_2 positive integers, you can think of " a^b_1" as meaning "a multiplied by itself b_1 times" so that (a^{b_1})^b_1 is "(a multiplied by itself b_1 times) multiplied by itself b_2 times" so that (a^{b_1})^{b_2} is a multiplied by itself a total of b_1b_2 times. Then, the standard procedure is to define define a^b for b negative, a fraction, and an irrational number so that (a^b_1)(a^b_2)= a^{b_1+ b_2}, (a^b_1)^b_2= a^{b_1b_2}, and so that a^b is a continuous function of b.
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  5. #5
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    Can I say the following:


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

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

    <br />
(a^{b_1})^{b_2}=a^{b_1\;b_2}.\\<br />
=\exp[b_2\log(a^{b_1})].\\<br />
=\exp[b_2\;b_1\log(a)].\\<br />
=a^{b_2\;b_1}.\\<br />
=a^{b_1\;b_2}.<br />

    -Michael
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