Subtracting Exponents
To understand subtraction of exponents, let's start from a simple example.
Example. Suppose, we want to find $$$\frac{{{{2}}^{{7}}}}{{{{2}}^{{4}}}}$$$.
We already learned about positive integer exponets, so we can write, that $$${{2}}^{{7}}={2}\cdot{2}\cdot{2}\cdot{2}\cdot{2}\cdot{2}\cdot{2}$$$ and $$${{2}}^{{4}}={2}\cdot{2}\cdot{2}\cdot{2}$$$.
So, $$$\frac{{{{2}}^{{7}}}}{{{{2}}^{{4}}}}=\frac{{{2}\cdot{2}\cdot{2}\cdot{\color{red}{{{2}\cdot{2}\cdot{2}\cdot{2}}}}}}{{{\color{red}{{{2}\cdot{2}\cdot{2}\cdot{2}}}}}}={2}\cdot{2}\cdot{2}={{2}}^{{3}}$$$.
Let's see what have we done. We counted number of 2's in $$${{2}}^{{7}}$$$, then counted number of 2's in $$${{2}}^{{4}}$$$. Since we divided, we canceled common terms. Note, that $$${7}-{4}={3}$$$.
It appears, that this rule works not only for positive integer exponents, it works for any exponent.
Rule for subtracting exponents: $$$\color{purple}{\frac{a^m}{a^n}=a^{m-n}}$$$.
Word of caution. It doesn't work, when bases are not equal.
For example, $$$\frac{{{{4}}^{{5}}}}{{{{3}}^{{2}}}}=\frac{{{4}\cdot{4}\cdot{4}\cdot{4}\cdot{4}}}{{{3}\cdot{3}}}$$$ which is neither $$${{4}}^{{3}}$$$ nor $$${{3}}^{{3}}$$$.
Word of caution. Above rule doesn't work for addition and subtraction.
For example, $$${{2}}^{{7}}-{{2}}^{{4}}\ne{{2}}^{{3}}$$$, because $$${{2}}^{{7}}-{{2}}^{{4}}={128}-{16}={112}$$$ and $$${{2}}^{{3}}={8}$$$. Clearly, $$${112}\ne{8}$$$.
Let's go through a couple of examples.
Example 2. Find $$$\frac{{{{2}}^{{3}}}}{{{{2}}^{{-{5}}}}}$$$.
It doesn't matter, that exponent is negative.
Just proceed as always: $$$\frac{{{{2}}^{{3}}}}{{{{2}}^{{-{5}}}}}={\left({{2}}^{{{3}-{\left(-{5}\right)}}}\right)}={{2}}^{{{3}+{5}}}={{2}}^{{8}}$$$.
Even when exponents are fractional, we use the same rule!
Example 3. Find $$$\frac{{{{3}}^{{\frac{{1}}{{4}}}}}}{{{{3}}^{{\frac{{2}}{{3}}}}}}$$$.
$$$\frac{{{{3}}^{{\frac{{1}}{{4}}}}}}{{{{3}}^{{\frac{{2}}{{3}}}}}}={{3}}^{{\frac{{1}}{{4}}-\frac{{2}}{{3}}}}={{3}}^{{-\frac{{5}}{{12}}}}=\frac{{1}}{{{3}}^{{\frac{{5}}{{12}}}}}$$$.
We can handle radicals, also, because radicals can be rewritten with the help of exponent.
Example 4. Rewrite, using positive exponent: $$$\frac{{\sqrt[{{8}}]{{{3}}}}}{{\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}}$$$.
First we rewrite numbers, using exponents and then apply the rule:
$$$\frac{{\sqrt[{{8}}]{{{3}}}}}{{\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}}=\frac{{{{3}}^{{\frac{{1}}{{8}}}}}}{{\sqrt[{{7}}]{{{{3}}^{{-{2}}}}}}}=\frac{{{{3}}^{{\frac{{1}}{{8}}}}}}{{{{3}}^{{-\frac{{2}}{{7}}}}}}={{3}}^{{\frac{{1}}{{8}}-{\left(-\frac{{2}}{{7}}\right)}}}={{3}}^{{\frac{{23}}{{56}}}}$$$.
Finally, we can see now, why $$${{a}}^{{0}}={1}$$$.
Indeed, $$${{a}}^{{0}}={{a}}^{{{n}-{n}}}=\frac{{{{a}}^{{n}}}}{{{{a}}^{{n}}}}={1}$$$.
Now, it is time to exercise.
Exercise 1. Find $$$\frac{{{{3}}^{{5}}}}{{{{3}}^{{2}}}}$$$.
Answer: $$${{3}}^{{3}}={27}$$$.
Exercise 2. Can we use rule for adding exponents to find $$$\frac{{{{5}}^{{5}}}}{{{{3}}^{{5}}}}$$$?
Answer: No, bases are not equal.
Exercise 3. Find $$$\frac{{{{4}}^{{\frac{{5}}{{3}}}}}}{{{{4}}^{{\frac{{2}}{{3}}}}}}$$$.
Answer: $$${4}$$$.
Exercise 4. Find $$$\frac{{{{3}}^{{2}}}}{{{{3}}^{{-\frac{{1}}{{5}}}}}}$$$.
Answer: $$${{3}}^{{{2}+\frac{{1}}{{5}}}}={{3}}^{{\frac{{11}}{{5}}}}={\sqrt[{{5}}]{{{{3}}^{{11}}}}}$$$.
Exercise 5. Find $$$\frac{{\sqrt[{{7}}]{{\frac{{1}}{{27}}}}}}{{\sqrt[{{8}}]{{{9}}}}}$$$.
Answer: $$$\frac{{\sqrt[{{7}}]{{{{3}}^{{-{3}}}}}}}{{\sqrt[{{8}}]{{{{3}}^{{2}}}}}}=\frac{{1}}{{{3}}^{{\frac{{19}}{{28}}}}}$$$.