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Solving for a variable

  1. Jan 27, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]{\frac {65-75\,{e^{-5\,k}}}{1-{e^{-5\,k}}}}={\frac {60-75\,{e^{-10\,k}
    }}{1-{e^{-10\,k}}}}[/tex]

    2. Relevant equations
    [tex]{e}^{x}=k[/tex] implies [tex]x=\ln \left( k \right)[/tex], as well as other properties of logarithms.

    3. The attempt at a solution
    Maple makes short work of this, giving [tex]k=1/5\,\ln \left( 2 \right) [/tex], but I'm totally lost as to how to solve it myself.
     
  2. jcsd
  3. Jan 28, 2008 #2
    I you set
    [tex]e^{-5\,k}=z[/tex]
    then what
    [tex]e^{-10\,k}[/tex]
    equals to?
     
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