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Math Help - Help evalutating numbers with exponents

  1. #1
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    Help evalutating numbers with exponents

    Hi can someone show me step by step how to evaluate these problems plz. Thanks in advance =]

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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by ferken View Post
    Hi can someone show me step by step how to evaluate these problems plz. Thanks in advance =]

    Okay, there are basically 3 rules to know :
    a^ba^c=a^{b+c}
    (a^b)^c=(a^c)^b=a^{bc}
    \frac{1}{a^b}=a^{-b}


    For the first one :

    \frac{32^{-\frac 15}}{(8^{\frac 53}-625^{\frac 12})^{-1}}=32^{-\frac 15} \left(8^{\frac 53}-625^{\frac 12}\right)

    Note that 32=2^5, 8=2^3, 625=5^4

    Hence 32^{-1/5}=(2^5)^{-1/5}=2^{-1}
    8^{5/3}=(2^3)^{5/3}=2^5
    625^{1/2}=(5^4)^{1/2}=5^2


    So your expression is now :

    =2^{-1} \left(2^5-5^2\right)=\frac 12 \left(32-25\right)=\frac 72
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  3. #3
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    Thanks for the help, but im still kinda stuck for the second one.
    Basically my problem is that i try to change it so that they all have the same base, but i am having troubles.
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  4. #4
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    Try to change them into common bases, in this case, 2 and 3. And then use the rules:

    Quote Originally Posted by Moo View Post
    Hello,

    Okay, there are basically 3 rules to know :
    a^ba^c=a^{b+c}
    (a^b)^c=(a^c)^b=a^{bc}
    \frac{1}{a^b}=a^{-b}
    and also (ab)^c=(a^c)(b^c)
    to combine liked terms.

    You may want to try something like this:

    \frac{(27^{\frac14})(3^{\frac12})(6^{\frac34})}{2^  {-\frac14}}
    =\frac{(3^{\frac34})(3^{\frac12})(2^{\frac34})(3^{  \frac34})}{2^{-\frac14}}

    and so forth.
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