Computing $4^{m+1} \cdot 9^{n-1}$ in terms of $2^m \cdot 3^n$

Question

So I got this math question that I have to do. Unfortunately I don't understand a thing. The question is:

If $2^m \cdot 3^n = a$, what is $4^{m+1} \cdot 9^{n-1}$?

I will be grateful for any and all help.

Answer 1

Notice, $$4^{m+1}\cdot 9^{n-1}$$ $$=(2^2)^{m+1}\cdot (3^2)^{n-1}$$ $$=(2)^{2m+2}\cdot (3)^{2n-2}$$ $$=2^{2m}\cdot 2^2\cdot 3^{2n}\cdot 3^{-2}$$ $$=\frac{4}{9}(2^{2m}\cdot 3^{2n})$$ $$=\frac{4}{9}(2^{m}\cdot 3^{n})^2$$ Setting $2^{m}\cdot 3^{n}=a$ $$=\frac{4}{9}(a)^2=\color{red}{\frac{4}{9}a^2}$$

Answer 2

$$4^{m+1}\cdot 9^{n-1}= 2^{2(m+1)} \cdot 3^{2(n-1)} = 2^{2m}\cdot 2^2 \cdot 3^{2n} \cdot 3^{-2} \\ = \frac{4}{9}(2^m \cdot 3^n)^2 = ... $$

I trust you can finish it off.