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Algebra problem

  1. Mar 22, 2009 #1

    I have the following equation:

    [tex]\Theta = log \left(\frac{\Pi}{1-\Pi}\right) [/tex]

    I want to re-arrange it for [tex]\Pi[/tex]

    Here's my attempt:

    [tex]\Theta = log \left( \frac{\Pi}{1-\Pi}\right) [/tex]

    [tex]\Theta = log \left( \Pi \right) - log \left(1-\Pi \right) [/tex]

    [tex]exp^{\Theta} = \Pi - (1-\Pi) [/tex]

    [tex]exp^{\Theta} = 2\Pi - 1 [/tex]

    [tex]1 + exp^{\Theta} = 2\Pi[/tex]

    [tex] \Pi = \frac{1 + exp^{\Theta}}{2} [/tex]

    The answer should be

    [tex] \Pi = \frac{exp^{\Theta}}{1+exp^{\Theta}} [/tex]

    Any idea where I'm going wrong?
  2. jcsd
  3. Mar 22, 2009 #2


    User Avatar
    Homework Helper

    Your mistake was on the 2nd step you made: [tex]exp^{\Theta} = \Pi - (1-\Pi) [/tex]

    Try using the basic definition of logs: [tex]log_ab=c[/tex] hence [tex]a^c=b[/tex]
    Last edited: Mar 22, 2009
  4. Mar 22, 2009 #3
    [tex]\theta =\log \left(\frac{\Pi }{1+\Pi }\right)[/tex]

    [tex]\frac{\Pi }{1+\Pi }=\exp (\theta )[/tex]

    [tex]\Pi =\exp (\theta )+\exp (\theta ) \Pi[/tex]

    [tex][1+\exp (\theta )]\Pi =\exp (\theta )[/tex]

    [tex]\Pi =\frac{\exp (\theta )}{1+\exp (\theta )}[/tex]
  5. Mar 22, 2009 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    This should be
    [tex]1- \exp(\theta)]\Pi= \exp(\theta)[/tex]

    [tex]\Pi =\frac{\exp (\theta )}{1-\exp (\theta )}[/tex]
    Last edited: Mar 22, 2009
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