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Math Help - find all real numbers x for logx=x/100

  1. #1
    Super Member bigwave's Avatar
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    find all real numbers x for logx=x/100

    Find (with an error of less than \frac{1}{10} ) all real numbers x for which \log_{10}x = \frac{x}{100}

    I assume this can be rewritten as x = 10^{\frac{x}{100} but don't know what next step would be

    the answers given are 1.0238 and 237.57 which I got with the TI89 but not sure the steps taken to derive it.

    thanks for help

    r
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  2. #2
    Senior Member Sambit's Avatar
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    The only way is to use the method of iteration. I made up the following code in R:

    [php]d=1.5;x=1.5
    while(d>.000001)
    {
    x.updated=100*log(x,base=10)
    d=abs(x.updated-x)
    x=x.updated
    }
    x.updated[/php]

    [php]
    d=1.5;x=1.5
    while(d>.000001)
    {
    x.updated=10^(x/100)
    d=abs(x.updated-x)
    x=x.updated
    }
    x.updated[/php]

    Run it in R; you will get the results.
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  3. #3
    Super Member bigwave's Avatar
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    Quote Originally Posted by Sambit View Post
    The only way is to use the method of iteration
    however this problem is out of book written in 1971 before writting a program for iteration was open to most students
    I would presume there would be another way to solve it.

    just curious
    r
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  4. #4
    MHF Contributor
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    you can deduce from the shape of the graph that there are at most 2 solutions. (the right hand side is a straight line, the LHS has an always decreasing gradient).

    I assume there is a clever way of getting the answer, but since i dont know it i would note that a talylor expansion of the LHS will give us a polynomial, which hopefully can be solved using conventional methods.

    Try a taylor expansion about x=1 of log_{10}x

    NB:
    f'(x) = x^{-1}log_{10} e
    f''(x) = -x^{-2}log_{10}e

    Taylor expansion:
    log_{10}x \approx log_{10}1 + (x-1)x^{-1}k -0.5x^{-2}(x-1)^2k

    where k= log_10(e)

    **thinking ahead, this is going to be a cubic. it's too late in the evening for that so ill use a 1 step taylor expansion instead

    log_{10}x \approx log_{10}1 + (x-1)x^{-1}k

    So solve
    log_{10}1 + (x-1)x^{-1}k = 0.01x

    0 + (x-1)x^{-1}k = 0.01x

     (x-1)k = 0.01x^2

    This is a quadratic in x, with solutions

    x=1.02 (pretty good!)
    x=42 (not so good!)

    presumably you could get closer by keeping more terms in the taylor expansion and solving a higher order polynomial.
    Last edited by SpringFan25; April 12th 2011 at 01:51 PM.
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  5. #5
    Master Of Puppets
    pickslides's Avatar
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    Quote Originally Posted by bigwave View Post
    however this problem is out of book written in 1971 before writting a program for iteration was open to most students
    I would presume there would be another way to solve it.

    just curious
    r
    Well how about an iterative method by hand? Bisection method comes to mind.
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  6. #6
    Senior Member Sambit's Avatar
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    Quote Originally Posted by bigwave View Post
    however this problem is out of book written in 1971 before writting a program for iteration was open to most students
    I would presume there would be another way to solve it.

    just curious
    r
    Of course you can do the method of iteration by hand. Method of bisection is a good idea, as pickslides said. I wrote that code for solving it with lesser labour and computation than by hand.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by bigwave View Post
    Find (with an error of less than \frac{1}{10} ) all real numbers x for which \log_{10}x = \frac{x}{100}

    I assume this can be rewritten as x = 10^{\frac{x}{100} but don't know what next step would be

    the answers given are 1.0238 and 237.57 which I got with the TI89 but not sure the steps taken to derive it.

    thanks for help

    r
    One step of Newton Raphson on f(x)=x-10^{x/100} with x_0=1 will do the job (possibly another step to confirm the error is small enough).

    CB
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  8. #8
    Super Member bigwave's Avatar
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    thanks everyone what i learned was quite valuable
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