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
$$1- \exp(\theta)]\Pi= \exp(\theta)$$

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

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