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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:
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!)
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.
The solutions here are very nice. I will consider this problem as a number theory problem. To do this, I will need to adopt Case 2 and Case 3 from @Lanier Freeman 's solution.
So, if we consider Case I only, we see that the number $n=3^a 2^b$ with $a, b\geq 0$ satisfies $$ 3\tau(n) = n $$ where $\tau(n)$ is the number of divisors of $n$. By an elementary inequality $$ \tau(n)\leq 2\sqrt n, $$ we have $$ n\leq 6\sqrt n. $$ Then, we need to consider only $n\leq 36$. Then exhaustive search will give us the answer.
In fact, by @BarryCipra 's observation, we have $3|n$, and this gives $$ 1\leq a \leq 3. $$ We take one by one,
If $a=1$, then $6(b+1)=3\cdot 2^b$. The solution is $(a,b)=(1,3)$.
If $a=2$, then $9(b+1) = 9 \cdot 2^b$. The solution is $(a,b)=(2,0)$ or $(2,1)$.
If $a=3$, then by $n\leq 36$, we have $b=0$. This cannot give a solution since $3\tau(n)=12\neq 27=n$.
