Properties of Exponents (Rules)

Properties (rules) of exponents:

Zero power: ${{a}}^{{0}}={1}$ , ${a}\ne{0}$
Zero base: ${{0}}^{{a}}={0}$ , ${a}\ne{0}$
${{0}}^{{0}}$ is undefined
${{1}}^{{a}}={1}$
Negative exponent: ${{a}}^{{-{b}}}=\frac{{1}}{{{a}}^{{b}}}$ , ${b}\ne{0}$
Nth root: ${{a}}^{{\frac{{1}}{{n}}}}={\sqrt[{{n}}]{{{a}}}}$ , ${n}\ne{0}$
Addition of exponents: ${{a}}^{{m}}\cdot{{a}}^{{n}}={{a}}^{{{m}+{n}}}$
Subtraction of exponents: $\frac{{{{a}}^{{m}}}}{{{{a}}^{{n}}}}={{a}}^{{{m}-{n}}}$ , ${a}\ne{0}$
Multiplication of exponents: ${{\left({{a}}^{{m}}\right)}}^{{n}}={{a}}^{{{m}\cdot{n}}}={{\left({{a}}^{{n}}\right)}}^{{m}}$
Division of exponents: ${\sqrt[{{n}}]{{{{a}}^{{m}}}}}={{a}}^{{\frac{{m}}{{n}}}}$ , ${n}\ne{0}$
${\sqrt[{{m}}]{{{{a}}^{{m}}}}}={a}$ , if ${m}$ is odd
${\sqrt[{{m}}]{{{{a}}^{{m}}}}}={\left|{a}\right|}$ , if ${m}$ is even
${\sqrt[{{n}}]{{{{a}}^{{m}}}}}={{\left({\sqrt[{{n}}]{{{a}}}}\right)}}^{{m}}$ (just pay attention to signs and check, whether number exists)
Power of a product: ${{a}}^{{n}}\cdot{{b}}^{{n}}={{\left({a}{b}\right)}}^{{n}}$
Power of a quotient: $\frac{{{{a}}^{{n}}}}{{{{b}}^{{n}}}}={{\left(\frac{{a}}{{b}}\right)}}^{{n}}$ , ${b}\ne{0}$

We already covered all rules earlier, except last two.

To understand last two properties, consider the following example.

Example. Find ${{2}}^{{3}}\cdot{{4}}^{{3}}$ .

Let's rewrite numbers: ${{\left({\color{red}{{{2}}}}\right)}}^{{3}}\cdot{{\left({\color{green}{{{4}}}}\right)}}^{{3}}={\color{red}{{{2}\cdot{2}\cdot{2}}}}\cdot{\color{green}{{{4}\cdot{4}\cdot{4}}}}$ .

Now, regroup: ${\color{red}{{{2}\cdot{2}\cdot{2}}}}\cdot{\left({\color{green}{{{4}\cdot{4}\cdot{4}}}}\right)}={\left({\color{red}{{{2}}}}\cdot{\color{green}{{{4}}}}\right)}\cdot{\left({\color{red}{{{2}}}}\cdot{\color{green}{{{4}}}}\right)}\cdot{\left({\color{red}{{{2}}}}\cdot{\color{green}{{{4}}}}\right)}={{\left({2}\cdot{4}\right)}}^{{3}}$ .

Note, that on the last step, we wrapped the product, using exponent.

This property is valid for any exponent, so:

Power of a product: ${\color{purple}{{{{a}}^{{n}}\cdot{{b}}^{{n}}={{\left({a}\cdot{b}\right)}}^{{n}}}}}$ .

Similarly, it can be shown that $\frac{{{a}}^{{n}}}{{{b}}^{{n}}}={{\left(\frac{{a}}{{b}}\right)}}^{{n}}$ .

Power of a quotient: ${\color{purple}{{\frac{{{a}}^{{n}}}{{{b}}^{{n}}}={{\left(\frac{{a}}{{b}}\right)}}^{{n}}}}}$ , ${b}\ne{0}$ .

We can combine above rules to simplify more complex examples.

Example 2. Find $\frac{{{6}}^{{4}}}{{{3}}^{{4}}}$ .

Using power of a quotient rule, we can write, that $\frac{{{6}}^{{4}}}{{{3}}^{{4}}}={{\left(\frac{{6}}{{3}}\right)}}^{{4}}={{2}}^{{4}}={16}$ .

Now, let's see how combination of rules works.

Example 3. Rewrite, using positive exponents: ${\sqrt[{{5}}]{{{{3}}^{{2}}\cdot{{3}}^{{5}}}}}$ .

We first apply rule for adding exponents: ${\sqrt[{{5}}]{{{{3}}^{{2}}\cdot{{3}}^{{5}}}}}={\sqrt[{{5}}]{{{{3}}^{{{2}+{5}}}}}}={\sqrt[{{5}}]{{{{3}}^{{7}}}}}$ .

Now, apply rule for dividing exponents: ${\sqrt[{{5}}]{{{{3}}^{{7}}}}}={{3}}^{{\frac{{7}}{{5}}}}$ .

So, ${\sqrt[{{5}}]{{{{3}}^{{2}}\cdot{{3}}^{{5}}}}}={\sqrt[{{5}}]{{{{3}}^{{7}}}}}$ .

Finally, let's see how to apply more than two rules.

Example 4. Rewrite, using positive exponents ${{\left(\frac{{{\sqrt[{{5}}]{{{3}}}}\cdot{\sqrt[{{5}}]{{{4}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}$ .

First, we rewrite using exponents: ${{\left(\frac{{{\sqrt[{{5}}]{{{3}}}}\cdot{\sqrt[{{5}}]{{{4}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}={{\left(\frac{{{{3}}^{{\frac{{1}}{{5}}}}\cdot{{4}}^{{\frac{{1}}{{5}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}$ .

Now, apply power of a product rule: ${{\left(\frac{{{{3}}^{{\frac{{1}}{{5}}}}\cdot{{4}}^{{\frac{{1}}{{5}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}={{\left(\frac{{{{\left({3}\cdot{4}\right)}}^{{\frac{{1}}{{5}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}={{\left(\frac{{{{12}}^{{\frac{{1}}{{5}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}$ .

Next, use rule for subtracting exponents: ${{\left(\frac{{{{12}}^{{\frac{{1}}{{5}}}}}}{{{12}}^{{3}}}\right)}}^{{3}}={{\left({{12}}^{{\frac{{1}}{{5}}-{3}}}\right)}}^{{3}}={{\left({{12}}^{{-\frac{{14}}{{5}}}}\right)}}^{{3}}$ .

Next, apply rule for multiplying exponents: ${{\left({{12}}^{{-\frac{{14}}{{5}}}}\right)}}^{{3}}={{12}}^{{-\frac{{14}}{{5}}\cdot{3}}}={{12}}^{{-\frac{{42}}{{5}}}}$ .

Finally, apply negative exponent rule: ${{12}}^{{-\frac{{42}}{{5}}}}=\frac{{1}}{{{12}}^{{\frac{{42}}{{5}}}}}$ .

Answer: $\frac{{1}}{{{12}}^{{\frac{{42}}{{5}}}}}$ .

Now, practice a little.

Exercise 1. Rewrite, using positive exponents: ${\sqrt[{{4}}]{{{2}\cdot{{2}}^{{5}}}}}$ .

Answer: ${{2}}^{{\frac{{3}}{{2}}}}$ .

Exercise 2. Rewrite, using positive exponents: ${{\left(\frac{{{2}}^{{3}}}{{{4}}^{{5}}}\cdot{\sqrt[{{5}}]{{{{5}}^{{7}}}}}\right)}}^{{0}}$ .

Answer: ${1}$ . Hint: as long as base is non-zero, raising to zero power gives 1.

Exercise 3. Rewrite, using positive exponents: ${\sqrt[{{5}}]{{{\sqrt[{{4}}]{{{{2}}^{{3}}\cdot{{6}}^{{3}}\cdot{{12}}^{{5}}}}}}}}$ .

Answer: ${{12}}^{{\frac{{2}}{{5}}}}$ .

Exercise 4. Simplify: ${{\left(\frac{{{{2}}^{{3}}}}{{{{6}}^{{3}}}}\cdot{{3}}^{{5}}\right)}}^{{7}}$ .

Answer: ${{3}}^{{14}}$ .

Exercise 5. Rewrite, using positive exponents: ${\sqrt[{{7}}]{{{{\left(\frac{{\sqrt[{{5}}]{{{{\left(-{2}\right)}}^{{7}}}}}}{{-{{2}}^{{3}}}}\cdot{{\left(-{2}\right)}}^{{7}}\right)}}^{{2}}}}}$ .

Answer: ${{2}}^{{\frac{{54}}{{35}}}}$ . Hint. pay attention to signs: ${{\left({{\left(-{2}\right)}}^{{\frac{{27}}{{5}}}}\right)}}^{{2}}={{2}}^{{\frac{{54}}{{5}}}}$ . Minus vanishes, because we square.