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Math Help - Difficult Power Rule task (without calculator)

  1. #1
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    Difficult Power Rule task (without calculator)

    I really need help with this one.

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  2. #2
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    2^7 = 128. So...
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  3. #3
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    Quote Originally Posted by undefined View Post
    2^7 = 128. So...
    How does that help him find an estimation for 128^48 without using a computer?
    128^{48}=(2^{14})^{24}=16384^{24}

    Or am I missing something? All I know is I wouldn't want to do 16384^14 by hand.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by 1005 View Post
    How does that help him find an estimation for 128^48 without using a computer?
    128^{48}=(2^{14})^{24}=16384^{24}

    Or am I missing something? All I know is I wouldn't want to do 16384^14 by hand.
    All we need is an estimate. Why not take

    128^{48}=16384^{24} \approx (10^4)^{24} = 10^{96}

    ?
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  5. #5
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    Hello, kevin3000!

    This is a strange problem . . .


    Calculate an approximate value to 128^{48}
    if you know that: 2^{14} = 16,\!384 . . I don't see how this is useful.

    I would do it like this . . .

    We have: . N \;=\;128^{48} \:=\:\left(2^7\right)^{48} \:=\:2^{336}


    Since 2^{10} \:=\:1024 \:\approx\:1000 \:=\:10^3

    we have: . N \;=\;2^6\cdot2^{330} \;=\;2^6\cdot (2^{10})^{33}

    . . . . . . . . N\;\approx\;2^6\cdot(10^3)^{33} \;=\;64\cdot10^{99}


    Therefore: . N \;\approx\;6.4 \times 10^{100}

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  6. #6
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    undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)

    and Soroban's approximation is 1 order of magnitude off with 54.28% error.
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  7. #7
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by 1005 View Post
    undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)

    and Soroban's approximation is 1 order of magnitude off with 54.28% error.
    It's funny yes but I was merely following the instructions of the problem (or trying to do so, since the problem doesn't make much sense); given freedom, I would also use the approximation 2^10 \approx 10^3.

    Anyway if you go high enough, 2^10 \approx 10^3 will give you 99.99% error as well.
    Last edited by undefined; July 20th 2010 at 03:21 PM.
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  8. #8
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    Estimates

    [QUOTE=1005;539205]undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)

    and Soroban's approximation is 1 order of magnitude off with 54.28% error.[/QUOTE




    Soroban's approx is not one magnitude off. It can be expressed as 0.64E101. Calculator answer is 1.4xE 101

    bjh
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