How do I express 108 as the product of powers of its prime factors? How do I express $108$ as the product of powers of its prime factors?
I've got $2^2 \times 3^3$ is this correct?
 A: First notice that $108$ is even so $2$ is a factor:
$$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow \\ 2 & &  & &54\end{matrix}$$
$54$ is also even:
$$\begin{matrix} & & 54 \\ & \large\swarrow & & \large\searrow \\ 2 & &  & &27\end{matrix}$$
$27$ is not even so let's check the next prime: $3$.  We can see that $3$ divides $27$ because $2+7=9$ is a multiple of $3$.  So:
$$\begin{matrix} & & 27 \\ & \large\swarrow & & \large\searrow \\ 3 & &  & &9\end{matrix}$$
And then hopefully you can see that $9$'s prime factorization is just $3\times 3$.  So putting all of these steps together we see that
$$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow  \\ \color{red}2 & & & & 54 \\ & & & \large\swarrow & & \large\searrow \\ & & \color{red}2 & & & & 27 \\ & & & & & \large\swarrow & & \large\searrow \\ & & & & \color{red}3 & & & & 9 \\ & & & & & & & \large\swarrow & & \large\searrow \\ & & & & & & \color{red}3 & & & & \color{red}3\end{matrix}$$
Thus 
$$108 = 2\times 2 \times 3 \times 3\times 3 = 2^2\times 3^3$$
A: Yes you are correct. 
Just divide 108 by its prime factors and count them !
$\frac{108}{2}=54, \frac{54}{2}=27 \Rightarrow$ $2^2$ 
$\frac{27}{3}=9, \frac{9}{3}=3 ,\frac{3}{3}=1 \Rightarrow 3^3$ 
So $108=2^2\cdot3^3$
A: Remember that multiplication is commutative and distributive. So $2^2 = 2 \times 2 = 4$ and $3^3 = 3 \times 3 \times 3 = 27$. Then when you do $4 \times 27 = 108$ what you "got" is confirmed as correct.
But if you're still unsure, try taking $2 \times 2 \times 3 \times 3 \times 3$ and redistributing the multiplicands, e.g., $6 \times 6 \times 3 = 108$ also. Note that that's half of $6 \times 6 \times 6 = 216 = 2^3 \times 3^3$. Mwahahahaha!
