REGRESSION :A log-linear interaction term How do I interpret a log-linear interaction term, is it possible? 
My model: 
$Y= B_1 + B_2\log X_1 + B_3X_2 + B_4(\log X_1 X_2) + u $
 A: Correct Model
As @whuber suggested, your interaction term should consist of the product of $\log X_1$ and $X_2$. That means your model should be specified as:
$Y= B_1 + B_2\log X_1 + B_3X_2 + B_4(\log X_1)(X_2) + u $
In this model, the effect of $\log X_1$ on $Y$ is assumed to depend on $X_2$. Conversely, the effect of $X_2$ on $Y$ is assumed to depend on $\log X_1$. 
What does the effect of of $\log X_1$ on $Y$ look like, given $X_2$?
To see how the effect of $\log X_1$ on $Y$ depends on $X_2$, you can re-write the model as follows:
$Y= B_1 + (B_2 + B_4X_2)\log X_1 + B_3X_2 + u $
From this new expression of the model, you can see that the effect of $\log X_1$ on $Y$ is quantified by the slope $B_2 + B_4X_2$. In particular, given $X_2$, every 1-unit increase in the value of $\log X_1$ is associated with a change of $B_2 + B_4X_2$ units in the expected value of $Y$. This change clearly depends on $X_2$. 
What does the effect of of $X_2$ on $Y$ look like, given $X_1$?
To see how the effect of $X_2$ on $Y$ depends on $\log X_1$, you can re-write the model as follows:
$Y= B_1 + B_2\log X_1 + (B_3 + B_4 \log X_1)X_2 + u $
From this alternative formulation of the original model, you can surmise that the effect of $X_2$ on $Y$ is quantified by the slope $B_3 + B_4 \log X_1$. Thus, given $X_1$, every 1-unit increase in the value of $X_2$ is associated with a change of $B_3 + B_4 \log X_1$ units in the expected value of $Y$. 
