Solving for \Pi: Unraveling the Logarithmic Equation

[roadworx]

Hi,

I have the following equation:

\Theta = log \left(\frac{\Pi}{1-\Pi}\right)

I want to re-arrange it for \Pi

Here's my attempt:

\Theta = log \left( \frac{\Pi}{1-\Pi}\right)

\Theta = log \left( \Pi \right) - log \left(1-\Pi \right)

exp^{\Theta} = \Pi - (1-\Pi)

exp^{\Theta} = 2\Pi - 1

1 + exp^{\Theta} = 2\Pi

\Pi = \frac{1 + exp^{\Theta}}{2}

The answer should be

\Pi = \frac{exp^{\Theta}}{1+exp^{\Theta}}

Any idea where I'm going wrong?

 

[Mentallic]

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

Try using the basic definition of logs: log_ab=c hence a^c=b

 

[zasdfgbnm]

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

Try using the basic definition of logs: log_ab=c hence a^c=b

\theta =\log \left(\frac{\Pi }{1+\Pi }\right)

\frac{\Pi }{1+\Pi }=\exp (\theta )

\Pi =\exp (\theta )+\exp (\theta ) \Pi

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

\Pi =\frac{\exp (\theta )}{1+\exp (\theta )}

 

[HallsofIvy]

\theta =\log \left(\frac{\Pi }{1+\Pi }\right)

\frac{\Pi }{1+\Pi }=\exp (\theta )

\Pi =\exp (\theta )+\exp (\theta ) \Pi

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

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

\Pi =\frac{\exp (\theta )}{1+\exp (\theta )}

\Pi =\frac{\exp (\theta )}{1-\exp (\theta )}