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Question on solving an equation involving logs

  1. May 20, 2012 #1
    1. The problem statement, all variables and given/known data
    (Here, by 'log' I mean natural logarithm)

    Solve for x:

    a.log(x) + b.log(1-x) = c

    for a, b and c constants


    2. Relevant equations



    3. The attempt at a solution
    Hi everyone,

    This is so embarrassing but this is really stumping me! I know how to do it if a=b:

    a.log(x) + a.log(1-x) = c

    log(x) + log(1-x) = c/a

    log(x(1-x)) = c/a

    x(1-x) = exp(c/a)

    and then you get a quadratic in x which you can solve.

    However, I can't work out how to solve it if a is not equal to b, as I can't see how to group up the logs nicely without ending up with horrible powers of x. What direction should I go in?

    Thanks for any help!
     
  2. jcsd
  3. May 20, 2012 #2

    SammyS

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    For this general case, I doubt that any approach will give a "nice" result.

    Divide both sides by b.

    Use properties of logarithms:

    a∙log(u) = log(ua)

    log(p)+log(q) = log(p∙q)
     
  4. May 20, 2012 #3

    sharks

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    Wouldn't it be simpler if you had just typed ##\ln## then? :smile:
    [tex]a\ln (x) + b\ln (1-x) = c
    \\\ln x^a+\ln (1-x)^b=c[/tex]
     
  5. May 20, 2012 #4
    I thought it might come out looking like an 'In'... :)

    So if I get it down to:

    p[itex]^{a}[/itex](1-p)[itex]^{b}[/itex] = e^c

    how can I solve for p then? Assuming the a and b are large-ish numbers like, say, 10 and 20.
     
  6. May 20, 2012 #5

    sharks

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    Since ##a\not = b##, i think you'll have to first expand ##(1-p)^b## using binomial theorem, and then multiply ##p^a## into that expansion.
     
  7. May 20, 2012 #6

    SammyS

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    Of course you could rewrite this as
    [itex]\displaystyle p^{\,a/b}(1-p)=e^{\,c/b}[/itex]​
     
  8. May 21, 2012 #7
    Thanks guys, I appreciate it! :)
     
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