I've been playing around with this problem for half an hour now and I don't believe this is solvable through elementary logarithmic means. Your equation is equivalent to this:
$$(a+1)(b+1)=3^{a-1}\cdot2^b$$
Case 1: Quadrant I & Axes
Now consider the following for nonnegative integer solutions:
$$3^{a-1}=a+1$$
$$2^b=b+1$$
The first equation is clearly true only for $a=1,a=2\;|\;a\in\Bbb Z$. Plugging either $a=1$ or $a=2$ into the original, we get the second equation. Now let's consider it:
Clearly it is only true for $b=3$ ($b\in\Bbb Z$) if $a=1$.
Clearly it is only true for $b=0$ and $b=1$ ($b\in\Bbb Z$) if $a=2$.
Ergo, three of our solutions for $a$ and $b$ are $(1,3)$, $(2,0)$, and $(2,1)$.
Case 2: Quadrants II and IV
If $a$ were positive and $b$ negative (or vice versa), the left-hand side would be negative and the right would be positive, so there are no solutions where $a>0, b<0\;\cup\;a<0,b>0$.
Case 3: Quadrant III
For negative factors, since $3^{a-1}\cdot2^b\notin\Bbb Z\;|\;-a,-b\in\Bbb N$ and $(a+1)(b+1)\in \Bbb Z\;|\;-a,-b\in\Bbb N$, there are no negative solutions.
If I have made a mistake or overlooked another angle please notify me.