Probability identities questions just a few simple identities/manipulations questions:
$\textbf{P}(A\cap B\cap C) = \textbf{P}(A|B\cap C)\textbf{P}(B| C)\textbf{P}(C)$ is this equivalent to the reverse? i.e: $\textbf{P}(A\cap B\cap C) = \textbf{P}(A)\textbf{P}(B|A)\textbf{P}(B\cap C|A)$? Are the variables here interchangeable? 
these few questions are regarding logarithmic operations on probabilities: do log operations on probabilities behave any different? i.e: $\log(\textbf{P}(A)\textbf{P}(B)) = \log(\textbf{P}(A)) + \log(\textbf{P}(B))$? or is there something I am missing? Also if we have two probabilities say $x$ and $y$, what would the range of possible values be for $\log(x/y)$?
Thank you and apologies for the unstructured questions.
 A: Yes, the terms are interchangeable, but you have to be careful how you do so.   The six valid permutations are:
$$\begin{array}\mathsf P(A\cap B\cap C) &=\mathsf P(A\mid B\cap C)~\mathsf P(B\mid C)~\mathsf P(C)\\&=\mathsf P(B\mid A\cap C)~\mathsf P(A\mid C)~\mathsf P(C)\\&=\mathsf P(C\mid A\cap B)~\mathsf P(A\mid B)~\mathsf P(B)\\&=\mathsf P(A\mid B\cap C)~\mathsf P(C\mid B)~\mathsf P(B)\\&=\mathsf P(B\mid A\cap C)~\mathsf P(C\mid A)~\mathsf P(A)\\&=\mathsf P(C\mid A\cap B)~\mathsf P(B\mid A)~\mathsf P(A)\end{array}$$

As long as neither probability equals zero, then $\lg(\mathsf P(A)\cdot\mathsf P(B))=\lg(\mathsf P(A))+\lg(\mathsf P(B))$ as normal.

So therefore, $\lg\mathsf P(A\cap B\cap C)=\lg\mathsf P(A\mid B\cap C)+\lg\mathsf P(B\mid C)+\lg\mathsf P(C)$
A: For the first question, it is not true.
$$P(A)P(B|A)P(B\cap C|A)=P(A \cap B) \frac{P(A \cap B \cap C)}{P(A)}$$
but in general, we have $\frac{P(A \cap B)}{P(A)}\le 1$. 
First question for logarithm is fine.
Assuming that $x$ and $y$ are positive, $\frac{x}{y}$ can be any positive number. Hence, the logarithm can take values from $(-\infty, \infty)$.
