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Math Help - n*2^n = 2^53 HOw to solve?

  1. #1
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    n*2^n = 2^53 HOw to solve?

    n*2^n = 2^53

    HOw to solve for n??

    thank you
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  2. #2
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    Quote Originally Posted by tea217 View Post
    n*2^n = 2^53

    HOw to solve for n??

    thank you
    Formula iteration will solve this with a bit of cunning:

    Rewrite the equation as:

    n=ln[(2^55)/n]/log(2)

    Then iterate:

    n(r+1)=ln[(2^55)/n(r)]/log(2)

    where n(r) is the r-th iterate.

    Then with n(0)=1, we get:

    n(1)=53,
    n(2)=47.2721,
    n(3)=47.4371,
    n(4)=47.4321
    n(5)=47.4322
    n(6)=47.4322

    With n=47.4322 the two sides of the original equation agree to 5 significant digits.

    RonL
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  3. #3
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    Quote Originally Posted by tea217 View Post
    n*2^n = 2^53

    HOw to solve for n??

    thank you
    This can also be solved with the LamberW function.

    If A=w*e^w, then W=LambertW(A).

    Your equation may be written:

    n* e^(ln(2) n)=2^53,

    or:

    [ln(2)*n] e^[ln(2)*n]=ln(2)*2^53,

    so:

    ln(2)*n=LambertW(ln(2)*2^53),

    or:

    n=LambertW(ln(2)*2^53)/ln(2).

    Now LambertW(ln(2)*2^53)~=32.8775, so n=32.8775/ln(2) ~=47.4322.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Formula iteration will solve this with a bit of cunning:

    Rewrite the equation as:

    n=ln[(2^55)/n]/log(2)

    Then iterate:

    n(r+1)=ln[(2^55)/n(r)]/log(2)

    where n(r) is the r-th iterate.

    Then with n(0)=1, we get:

    n(1)=53,
    n(2)=47.2721,
    n(3)=47.4371,
    n(4)=47.4321
    n(5)=47.4322
    n(6)=47.4322

    With n=47.4322 the two sides of the original equation agree to 5 significant digits.

    RonL

    "n=ln[(2^55)/n]/log(2)"

    how did u get this statement?
    why is there a ln and log? what base is log(2) ?

    this method seems easier to understand than the other one, which i have never used that function before.

    But i really appricate your help.
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  5. #5
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    that lambert w function is incredible
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  6. #6
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    Quote Originally Posted by tea217 View Post
    "n=ln[(2^55)/n]/log(2)"

    how did u get this statement?
    why is there a ln and log? what base is log(2) ?

    this method seems easier to understand than the other one, which i have never used that function before.

    But i really appricate your help.
    Typo, default base for log is e, so when I forget to type ln and type log
    I usually mean log base e.

    We have:

    n 2^n = 2^53

    which we may rewrite as:

    n e^(n ln(2))=2^53

    so:

    e^(n ln(2))=(2^53)/n

    Now taking logs (ln's) gives:

    n ln(2)= ln((2^53)/n)

    n= ln((2^53)/n)/ln(2)

    The equation is reorganised in this manner so that formula iteration
    will converge (with the obvious rearrangement iteration diverges)

    RonL
    Last edited by CaptainBlack; February 19th 2007 at 03:28 AM.
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