1
$\begingroup$

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 $b$.

$\endgroup$
  • 3
    $\begingroup$ 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
4
$\begingroup$

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!)

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.