Simplifying $3^{-4}$ to $1/81$ How would you solve for $3^{-4}$? I know the answer is $1/81$ but I can't work out how you get there with this one.

Let's consider the expression
$$\frac{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3}$$
There are three more $3$s multiplied together on the top than there are on the bottom, and we write it more concisely as $3^3$.
Suppose now that the number of $3$s on top are equal to the number of threes on the bottom. In this case we would write $3^0$ (which simplifies to  just $1$).
$$\frac{3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3}$$
Following this model, we would expect that $3$ raised to a negative power would represent a fraction where the number of $3$s on the bottom exceed the number of $3$s on the top, as follows.
$$\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}$$which we would represent as $3^{-4}$.
Continuing we have$$\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}=\frac{3\cdot 3}{3\cdot 3\cdot ((3\cdot 3)\cdot (3\cdot 3))}=\frac{3\cdot 3}{3\cdot 3\cdot (9\cdot 9)}=\frac{1}{81}$$
