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 $b$.

  • 4
    $\begingroup$ Hint: Pretty soon, $2^b\gt b+1$ and $3^a\gt 3(a+1)$. Logarithms will not be needed. $\endgroup$ – André Nicolas May 26 '16 at 1:19

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:

  1. $a=1$ and $6(b+1)=3\cdot2^b$
  2. $b=0$ and $3(a+1)=3^a$
  3. $b=1$ and $6(a+1)=3^a\cdot2$
  4. $b=2$ and $9(a+1)=3^a\cdot4$
  5. $b=3$ and $12(a+1)=3^a\cdot8$

Tackling them one at a time....

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

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

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

  4. $3^a\gt{9\over4}(a+1)$ if $a\gt1$, but $9(1+1)\not=3\cdot4$.

  5. $3^a\gt{12\over8}(a+1)={3\over2}(a+1)$ 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!)

  • 1
    $\begingroup$ Not at all clunky. The beginning analysis shows there are only a small number of "live" cases, and now apart from a little computation it's over. $\endgroup$ – André Nicolas May 26 '16 at 4:36
  • $\begingroup$ @AndréNicolas, thanks, that describes the overall strategry to a tee. It still feels a bit clunky, though, to wind up saying pretty much the same thing over and over and over again. $\endgroup$ – Barry Cipra May 26 '16 at 12:20

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:


Case 1: Quadrant I & Axes

Now consider the following for nonnegative integer solutions:



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 $3\tau(n)=12\neq 27=n$.

  • $\begingroup$ Really nice use of the inequality on $\tau$. I like this answer better than my own. $\endgroup$ – Barry Cipra May 26 '16 at 13:17
  • $\begingroup$ Incidentally, I think it's easier to rule out negative values of $a$ or $b$ by pointing out that the left hand side is always an integer whereas the right hand side acquires a denominator if either variable is negative. (I didn't even mention this in my answer, though maybe I should edit it in.) $\endgroup$ – Barry Cipra May 26 '16 at 13:17

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