# 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: $\frac{{\sqrt[{{7}}]{{{{3}}^{{-{3}}}}}}}{{\sqrt[{{8}}]{{{{3}}^{{2}}}}}}=\frac{{1}}{{{3}}^{{\frac{{19}}{{28}}}}}$.