How to solve $3(a+1)(b+1)=3^a \times 2^b$? Hi I'm new to logarithms and not sure how to solve equations involving logarithms. I managed to find this equation to answer a problem solving question, however now I do not know how to solve the logarithm? Any guidance tips or help would be greatly appreciated.

$3(a+1)(b+1)=3^a \times 2^b$
Find all integer solutions for $a$ and $b$.


Note that $a+1\lt3^a$ if $a\gt1$ and $3(b+1)\lt2^b$ if $b\gt3$.  Consequently $3(a+1)(b+1)\lt3^a\cdot2^b$ unless either $a=1$ or $0\le b\le3$.  (Note, $a$ cannot be $0$, since the left hand side is divisible by $3$.)  Thus we have five subcases to consider:


*

*$a=1$ and $6(b+1)=3\cdot2^b$

*$b=0$ and $3(a+1)=3^a$

*$b=1$ and $6(a+1)=3^a\cdot2$

*$b=2$ and $9(a+1)=3^a\cdot4$

*$b=3$ and $12(a+1)=3^a\cdot8$


Tackling them one at a time....


*

*$2^b\gt2(b+1)$ if $b\gt3$, but the equation $2^b=2(b+1)$ is not solved by $b=0$, $1$, $2$, or $3$.

*$3^a\gt3(a+1)$ if $a\gt2$, $3^1\not=3(1+1)$, but $3^2=3(2+1)$, so $(a,b)=(2,0)$ is a solution.

*Same as 2.  $(a,b)=(2,1)$ is a solution.

*$3^a\gt{9\over4}(a+1)$ if $a\gt1$, but $9(1+1)\not=3\cdot4$.

*$3^a\gt{12\over8}(a+1)={3\over2}(a+1)$ if $a\gt1$, but $12(1+1)=3^1\cdot8$, so $(a,b)=(1,3)$ is a solution.
And that's all.  The equation $3(a+1)(b+1)=3^a\cdot2^b$ has exactly three solutions:  $(a,b)=(2,0)$, $(2,1)$, and $(1,3)$.
Remark:  This analysis feels a little clunky, but I don't see any simple way to streamline it.  Maybe someone else can. (Update:  i707107's $\tau$-based answer streamlines things considerably.  I wish I'd thought of it!)
