Bernoulli Logarithm

Question

How does this:

$$\log[\phi^y(1-\phi)^{1-y}]$$

Become this:

$$\log\left(\frac\phi{1-\phi}\right)y + \log(1-\phi)$$

Answer 1

Beginning with $$\log(\varphi^{y} \ (1-\varphi)^{1-y})$$

Since $\log a^x = x \log a$ and $\log a b = \log a + \log b$ we can rewrite the equation above as

$$y \log\varphi + (1-y) \log (1-\varphi)$$

Multiplying the right hand side of $(1-y) \log (1-\varphi)$ we have

$$y \log\varphi + \log (1-\varphi)-y \log (1-\varphi)$$

Factoring out $y$ we have

$$y (\log\varphi -\log (1-\varphi))+ \log (1-\varphi)$$

Since $\log a - \log b = \log \dfrac {a}{b}$ we get

$$y (\log (\dfrac {\varphi}{\varphi-1})) + \log (1-\varphi)$$