Why is the decision function for probabilistic models a quotient (when we only consider two models)? Take for example, that we want to find the probabilistic model for only two document types (doc can be + or -).
I was trying to understand why the way that we classify a document model was with the following decision rule (under the assumption equal probability of the two type of models):
$$log{\frac{P(D|\theta^+)}{P(D|\theta^-)}} =
\begin{cases}
\geq0, & \text{then +} \\
<0, & \text{then -}
\end{cases}
$$
Why didn't we for example, choose to be the deciding expression:
$$\frac{logP(D|\theta^+)}{logP(D|\theta^-)}$$
I feel it has something to do that what we originally had in mind to use:
$$\frac{P(D|\theta^+)}{P(D|\theta^-)}$$
but probably using logs made is more mathematically convenient or something.
But my point of my question is, why do we use the first decision rule that I posted? What is the intuition? Is the reason we are using it is for mathematical convenience or some deeper reason? What is wrong with the other two suggestion I have posted?

Notation:
D = some document (just a bag of words for the document).
+,- =  two different document classes.
$\theta^+$ = the parameter/model we have for the document class +.
$\theta^-$ = the parameter/model we have for the document class -.
 A: In your top expression, you're taking the log of the Bayes factor, the ratio of likelihoods of the observed data under two candidate models. A high Bayes factor favors the model in the numerator of the expression, and two models that produce equal likelihoods for the observed data will produce a Bayes factor equal to one.
Since $log(1) = 0$, the decision rule above, and this one, are equivalent:
$\frac{P(D|\theta^+)}{P(D|\theta^-)} =
\begin{cases}
\geq0, & \text{then +} \\
<0, & \text{then -}
\end{cases}$
But the middle expression will not produce equivalent results, e.g.:
$log\frac{e}{2} = 0.3068528 \ne 1.442695 = \frac{log(e)}{log(2)}$
Thus, we can't use the second. But, since $log\frac{a}{b} = log(a) - log(b)$, you can use the difference of the log likelihoods in the above decision rule with the same result.
If the likelihood functions include an exponential, then taking the log essentially lets you skip a computational step. Meaning, instead of computing $\frac{e^{\lambda+}}{e^{\lambda-}}$ and comparing to 1, we can just compute $\lambda^+ - \lambda^-$ and compare to 0. In other words, we can express the decision rule in a more computationally efficient, but equivalent, form. In models in which the likelihood is a product of exponentials, this could save many computations.
