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Thread: evaluation of exponential expression

  1. #1
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    evaluation of exponential expression

    without a calculator, how do i determine that :
    32^3/5
    = 8
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  2. #2
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    Quote Originally Posted by dymand68 View Post
    without a calculator, how do i determine that :
    32^3/5
    = 8
    $\displaystyle (32^{1/5})^3$ and you're expected to know what $\displaystyle 32^{1/5}$ is.
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  3. #3
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    Quote Originally Posted by dymand68 View Post
    without a calculator, how do i determine that :
    32^3/5
    = 8

    Definitions: $\displaystyle 32^{3\slash 5} = \left(32^{1\slash 5}\right)^3=\left(\sqrt[5]{32}\right)^3$ . Now just check that $\displaystyle 2^5 = 32$ so...

    Tonio
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  4. #4
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    the problem is:
    evaluate:
    32^-3/5
    =1/32^3/5
    =???
    1/8 ?
    thus my original question: without a calculator, how do i Know that
    32^3/5
    = 8
    ...okay, find the fifth root of 32 and cube it, gotcha
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  5. #5
    Member integral's Avatar
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    If you want to be fancy

    This does not require a calculator at all!

    $\displaystyle 32^{3/5}=n$
    (take the log of both sides)

    $\displaystyle \textrm{log}_{32}32^{\frac{3}{5}}=\textrm{log}_{32 }8$

    $\displaystyle \textrm{log}_a(a)^x=x\therefore$

    $\displaystyle \frac{3}{5}=\textrm{log}_{32}8$
    log base change formula. (changes and base to two base 10 logs):
    $\displaystyle \textrm{log}_ab=\frac{\textrm{log}(b)}{\textrm{log }(a)} \therefore$

    $\displaystyle \frac{3}{5}=\frac{\textrm{log}8}{\textrm{log}32}$

    $\displaystyle \textrm{log}(a)^x=x\textrm{log}a$
    and
    $\displaystyle \textrm{log}(ab)=\textrm{log}(a)+\textrm{log}(b)$

    $\displaystyle \frac{3}{5}=\frac{3\textrm{log}2}{\textrm{log}4+\t extrm{log}8}$

    $\displaystyle \frac{3}{5}=\frac{3\textrm{log}2}{2\textrm{log}2+3 \textrm{log}2}$

    $\displaystyle \frac{3}{5}=\frac{3\textrm{log}2}{5\textrm{log}2}$

    $\displaystyle \frac{3}{5}=\frac{3}{5}$

    $\displaystyle \therefore 32^{3/5}=8$


    ((This is my 100th post! Whooo! ))
    Last edited by integral; Apr 1st 2010 at 10:35 PM.
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