Double product of $2^i\cdot 3^j$

Question

Self learning Trying to find a closed form for $\prod\limits_{i=1}^n\prod\limits_{j=1}^n 2^i3^j$ can't find any help on how to tackle this type of problem. Everything I saw on here went over my head. This is how I'm trying to solve it.

$$\prod\limits_{i=1}^n\ln\left[\prod\limits_{j=1}^n 2^i3^j\right]=\prod\limits_{i=1}^n\sum\limits_{j=1}^n\ln\left(2^i3^j\right)$$ $$ \begin{align*} \sum\limits_{j=1}^n\ln\left(2^i3^j\right) &= \sum\limits_{j=1}^n\ln(2^i)+\sum\limits_{j=1}^n\ln(3^j)\\ &=n\ln(2^i)+\sum\limits_{j=1}^njln(3)\\ &=n\ln(2^i)+ln(3)\sum\limits_{j=1}^nj\\ &=n\ln(2^i)+ln(3)\frac{n(n+1)}{2} \end{align*} $$ raise everything to e and simplify and I think you get $$\prod\limits_{i=1}^n\prod\limits_{j=1}^n 2^i3^j=\prod\limits_{i=1}^n\left[2^{i^n}+3^\frac{n(n+1)}{2}\right]$$ so gonna do the ln trick again. I also have no idea if this is close at all. hopefully it is because I'm not good a LaTex, and this might be a waste of time. $$ \begin{align*}\ln\left[\prod\limits_{i=1}^n\left[2^{i^n}+3^\frac{n(n+1)}{2}\right]\right]&=\sum\limits_{i=1}^n\ln\left[2^{i^n}+3^\frac{n(n+1)}{2}\right]\\ &=\sum\limits_{i=1}^nln(2^{i^n})ln\left(3^\frac{n(n+1)}{2}\right) \end{align*} $$

Now I don't know how to deal with the last sum. I don't think I can pull out that second ln to the front because there are n's in it. I also have a suspicion that there's an easy trick to make this simple.

Answer 1

The inner product doesn't depend on the index of the outer product, so you can move stuff around more liberally: $$ \prod\limits_{i=1}^n\left(\prod\limits_{j=1}^n 2^i3^j\right) = \prod\limits_{i=1}^n \left(2^{ni} \prod\limits_{j=1}^n 3^j \right) = \prod\limits_{i=1}^n 2^{ni} \left(\prod\limits_{j=1}^n 3^j \right)^n = 2^{n\frac{n(n+1)}{2}} 3^{n\frac{n(n+1)}{2}} = \left(2^n\cdot 3^n\right)^\frac{n(n+1)}{2} $$

Answer 2

You are on the right track, but you should place the log on the outside.

$$\log\prod\limits_{i=1}^n\prod\limits_{j=1}^n2^i3^j=\sum\limits_{i=1}^n\sum\limits_{j=1}^n(i\log 2+j\log 3)=\sum\limits_{i=1}^n\left(in\log2+\frac{n(n+1)}2\log3\right)$$$$=\frac{n^2(n+1)}2\cdot(\log2+\log3)=\frac{n^2(n+1)}2\cdot\log6$$Thus, our answer is $6^{\frac{n^2(n+1)}2}=(\sqrt{6})^{n^2(n+1)}$

Answer 3

There are quite a few ways to solve this. There's the log method, which you started on, which converts products into sums. But if you you can transfer your understanding of how sums work to the corresponding phenomena for products, you can work with the products directly. If you had $\sum\limits_{j=1}^n (2i+3j)$, you could split that up into $\sum\limits_{j=1}^n 2i+\sum\limits_{j=1}^n 3j$, and since the summand for $\sum\limits_{j=1}^n 2i$ doesn't depend on $j$, it turns into a product: $\sum\limits_{j=1}^n 2i = 2in$, which leaves us with $2in+\sum\limits_{j=1}^n 3j$. You reached a similar result with logs, but if you understand the correspondence between sums and products, you can directly see that just as $\sum\limits_{j=1}^n (2i+3j) = 2in+\sum\limits_{j=1}^n 3j$, we also have $\prod\limits_{j=1}^n (2^i3^j) = 2^{in}\prod\limits_{j=1}^n 3^j$. You can see that going from sums to products, I converted all addition to multiplication, and all multiplication to exponentiation. And since multiplication corresponds to addition within the exponent, we can take $\prod\limits_{j=1}^n 3^j$ and convert it to $3^{\sum\limits_{j=1}^nj}$, or $3^{\frac{n(n+1)}2}$. So the total expression is $\prod\limits_{i=1}^n\prod\limits_{j=1}^n (2^i3^j)=\prod\limits_{i=1}^n2^{ni}3^{\frac{n(n+1)}2}=\prod\limits_{i=1}^n2^{ni}\prod\limits_{i=1}^n3^{\frac{n(n+1)}2}=\prod\limits_{i=1}^n3^{\frac{n(n+1)}2}\prod\limits_{i=1}^n2^{ni}=\left(3^{\frac{n(n+1)}2}\right)^n\prod\limits_{i=1}^n2^{ni} = 3^{\frac{n^2(n+1)}2} \prod\limits_{i=1}^n2^{ni} $.

We can then use a similar process (or symmetry) to get $3^{\frac{n^2(n+1)}2} \prod\limits_{i=1}^n2^{ni} =3^{\frac{n^2(n+1)}2} 2^{\frac{n^2(n+1)}2} = 6^{\frac{n^2(n+1)}2} $.