$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x\log \pi + (n-x)\log(1-\pi)\;\;?$

Question

$$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x \log \pi + (n-x)\log(1-\pi)$$ this is what i have.

i dont understand how $\binom{n}{x}$ disappears, but the rest is fine.

I tried this, but it is getting complicated. Is there some way I can do things like this quickly and at the first glance?

$$\begin{align}\log\left(\binom{n}{x}\right) & = \log\left(\dfrac{n!}{x!(n-x)!}\right)\\ \\ & =\log(n!)-\log\left[x!(n-x)!\right] \\ \\ & = \log(n!)-\left[\log(x!)+\log(n-x)!\right] \\ \\ & = \;\; \cdots \end{align}$$

Answer 1

This is only true when $x=0$ or $x=n$. Otherwise the final term does not disappear.

Answer 2

so we use the property

$\log abc=\log a+\log b+\log c$

$\log(\left(\begin{matrix}n \\ x\end{matrix}\right)\pi^x(1-\pi)^{n-x})$

$=\log\left(\begin{matrix}n \\ x\end{matrix}\right)+\log \pi^x+\log(1-\pi)^{n-x}$

$=\log\left(\begin{matrix}n \\ x\end{matrix}\right)+x\log \pi+(n-x)\log(1-\pi)$

$\log\left(\begin{matrix}n \\ x\end{matrix}\right)=\log \Gamma(n+1)-\log\Gamma(x+1)-\log\Gamma(n-x+1)$

Answer 3

$\binom{n}{x}$ does not disappear and the identity does not hold. Try $n=4, x=3, \pi=0.5$, say, to get $\log(1/4)=2\log(1/2)$ on the left and $4\log(1/2)$ on the right.