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Minimum value of logarithmic equation

  1. Jun 29, 2008 #1
    1. The problem statement, all variables and given/known data

    For x>1, find the minimum possible value of [tex]2 log_{2}x-log_{x}(0.01)[/tex]

    3. The attempt at a solution
    Greater the value of x, greater is the value of expression. Right?
    I tried to differentiate it, but it was no help. The derivative becomes zero when |log x|=[tex]\sqrt{log 2}[/tex]

    Help me further.
     
  2. jcsd
  3. Jun 29, 2008 #2

    Defennder

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    Re: Log

    What did you get when you differentiate it?
     
  4. Jun 29, 2008 #3
    Re: Log

    Why do ya ask that? Is it wrong?
    I got:
    2*(1/x ln 2)-2*(1/log^{2} x)*(1/(x ln 10))
     
  5. Jun 29, 2008 #4

    Defennder

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    Re: Log

    Well, apparently I can't tell if it's correct unless I know your working. I can't read what you wrote that. Is it [tex]\frac{2 \ln x}{\ln 2} - \frac{2}{\log_2 x} \left( \frac{1}{x \ln 10} \right) [/tex].

    If so then I don't think it's correct.
     
  6. Jun 30, 2008 #5
    Re: Log

    [tex]\frac{2}{x ln 2}-\frac{2}{(log^{2} x)*(x ln 10)}[/tex]
     
  7. Jun 30, 2008 #6

    dynamicsolo

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    Re: Log

    This problem sets a similar trap to that of another problem you asked about. It would make life easier if you rewrote the [tex]log_{x}(0.01)[/tex] term as a log-base-2 term first, so you could combine it with the first term...

    (You could grind through the differentiation you have, but it is a rather cumbersome "hammer" to use on the problem.)
     
  8. Jun 30, 2008 #7
    Re: Log

    I got:
    [tex]\frac{2}{x ln 2}-\frac{2 ln 10}{x ln^2 x}[/tex]
    for the derivative.
    I set it equal to zero, cross multiplied and came up with:

    [tex]ln^2x=ln 10 ln 2[/tex]

    Dunno if that helps...your derivative was different than mine.
    CC
     
    Last edited: Jun 30, 2008
  9. Jun 30, 2008 #8
    Re: Log

    rewriting:
    [tex]2log_{2}x+\frac{2(1+log_{2}5)}{log_{2}x}[/tex]
     
  10. Jun 30, 2008 #9
    Re: Log

    Take the square root of both sides, ignore the absolute value bars, because for x>1 the thing is positive, then take the exponential of both sides. I got x=3.53722.......
    plug that back in to get the y value.
    CC
     
    Last edited: Jun 30, 2008
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