why $\exp[ \log (1-\mu)]= \exp[- \log(1 + \exp(\theta))]$

Question

why?

Let $\theta = \log\frac{\mu}{1-\mu}$ $$\begin{align*} &\exp[ \log (1-\mu)] \\ &= \exp[- \log(1 + \exp(\theta))] \\ \end{align*}$$

My calculate:

$$\begin{align*} &\exp[ \log (1-\mu)] \\ &= \exp[-\log \mu +\log(1-\mu) + \log\mu ] \\ &= \exp[\log\frac{1 - \mu}{\mu} + \log\mu ] \\ &= \exp[-\log\frac{\mu}{1- \mu} + \log\mu ] \text{ ...stucked in here}\\ & ...\\ &= \exp[- \log(1 + \exp(\theta))] \\ \end{align*}$$

I know $\exp(\log(x))=x$

Answer 1

You have: $$exp(-log {\mu \over 1-\mu}+log \mu)$$ Notice that $log(x) = -log(\frac 1x)$ and that $log(x)+log(y) = log(xy)$

So we have: $$exp(-log {\mu \over 1-\mu}+log \mu)=exp(-log {\mu \over 1-\mu}-log\frac1\mu)=exp(-log \frac 1 {1-\mu})$$

But now notice that we defined $\theta = log {\mu \over 1-\mu}=log{\mu-1+1 \over 1-\mu}=log({\mu-1 \over 1-\mu}+\frac 1 {1-\mu})=log(-1+\frac1 {1-\mu})$

Which means $exp(\theta) = -1+\frac1{1-\mu}$ and our $exp(-log \frac 1 {1-\mu})$ becomes $exp(-log (1+exp(\theta))$

You could really do this a lot quicker if you use the fact that $exp(-log(x))=\frac 1 x$ so that you can go straight from $1-\mu$ to $exp(-log \frac 1 {1-\mu})$ but I tried to stick to your solution.