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Is it possible to get an explicit solution for this?

  1. Dec 10, 2012 #1
    x = 10logx + 30 (log is log base 10)
    I cannot get to anything other than this implicit solution. By trial and error I can tell that x must be slightly more than 1/1000 but I would like to get an exact answer.
     
  2. jcsd
  3. Dec 11, 2012 #2

    lurflurf

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    Homework Helper

    You need a function such as the Lambert W function to get an explicit solution. Real solution are approximately .0001 and 46.6925.
     
  4. Dec 11, 2012 #3
    Thanks for the reply. I know you probably don't want to give me the answer right out but in looking at that Wikipedia article I still don't see how to do it. I am not able to get both x's on the same side in any manner resembling the examples from the article.
     
  5. Dec 11, 2012 #4

    lurflurf

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    It can be a little tricky. These log's are base e.

    x = (10/log(10))log(x) + 30
    (-log(10)/10)x=-log(x)+30(-log(10)/10)
    (-log(10)/10)x+log(x)=log(10^-3)
    (-log(10)/10)x+log((-log(10)/10)x)-log(-log(10)/10)=log(10^-3)
    (-log(10)/10)x+log((-log(10)/10)x)=log(-log(10)/10^4)
    (-log(10)/10)x=W(-log(10)/10^4)
    x=(-10/log(10))W(-log(10)/10^4)

    There are log's of negative numbers in there.


    For finding numerical answers, you can improve your guess systematically
    guess 50
    x = 10log10(x) + 30
    guess 0
    x=10^(x/10-3)

    Then in each case put the guess into the right hand side over and over until it changes very little.
    ie
    50
    10log10(50) + 30~46.9897000
    10log10(46.9897000) + 30~46.7200267
    ~46.6950
    and so forth
     
  6. Dec 11, 2012 #5

    uart

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    Science Advisor

    First convert your log to base "e", and then write your equation in the form,

    [tex]\ln(x) = ax + b[/tex]

    Now exponentiate both sided and put it in the form,

    [tex]x = e^b \, e^{ax}[/tex]

    Finally mult both sides by (-a) and rearrange into the form,

    [tex] (-ax) e^{-ax} = k[/tex]

    It should then be straightforward to use the Lambert W function.
     
    Last edited: Dec 11, 2012
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