# Multiplying Exponents

To understand multiplication of exponents, let's start from a simple example.

Example. Suppose, we want to find ${{\left({{2}}^{{3}}\right)}}^{{4}}$.

We already learned about positive integer exponets, so we can rewrite outer exponent: ${{\left({{2}}^{{3}}\right)}}^{{4}}={{2}}^{{3}}\cdot{{2}}^{{3}}\cdot{{2}}^{{3}}\cdot{{2}}^{{3}}$.

Now, using addition of exponents, we have that ${{\left({{2}}^{{3}}\right)}}^{{4}}={{2}}^{{3}}\cdot{{2}}^{{3}}\cdot{{2}}^{{3}}\cdot{{2}}^{{3}}={{2}}^{{{3}+{3}+{3}+{3}}}={{2}}^{{{3}\cdot{4}}}={{2}}^{{12}}$.

Let's see what have we done. We rewrote outer exponent, and then applied the rule for adding exponents.

But notice, that we added 3 four times, In other words we multiplied 3 by 4. Note, that ${3}\cdot{4}={12}$.

It appears, that this rule works not only for positive integer exponents, it works for any exponent.

Rule for subtracting exponents: $\color{purple}{\left(a^m\right)^n=a^{m\cdot n}}$.

Note. Since ${m}\cdot{n}={n}\cdot{m}$, then ${{\left({{a}}^{{m}}\right)}}^{{n}}={{\left({{a}}^{{n}}\right)}}^{{m}}$.

Example 2. Find ${{\left({{2}}^{{3}}\right)}}^{{-{15}}}$.

It doesn't matter, that exponent is negative.

Just proceed as always: ${{\left({{2}}^{{3}}\right)}}^{{-{15}}}={{2}}^{{{3}\cdot{\left(-{15}\right)}}}={{2}}^{{-{45}}}=\frac{{1}}{{{2}}^{{45}}}$.

Even when exponents are fractional, we use the same rule!

Example 3. Find ${{\left({{3}}^{{\frac{{1}}{{4}}}}\right)}}^{{{2}}}$.

${{\left({{3}}^{{\frac{{1}}{{4}}}}\right)}}^{{2}}={{3}}^{{\frac{{1}}{{4}}\cdot{2}}}={{3}}^{{\frac{{1}}{{2}}}}=\sqrt{{{3}}}$.

We can handle radicals, also, because radicals can be rewritten with the help of exponent.

Example 4. Rewrite, using positive exponent: ${{\left({\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}\right)}}^{{5}}$.

First we rewrite number, using exponents and then apply the rule:

${{\left({\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}\right)}}^{{5}}={{\left({\sqrt[{{7}}]{{{{3}}^{{-{2}}}}}}\right)}}^{{5}}={{\left({{3}}^{{-\frac{{2}}{{7}}}}\right)}}^{{5}}={{3}}^{{-\frac{{2}}{{7}}\cdot{5}}}={{3}}^{{-\frac{{10}}{{7}}}}=\frac{{1}}{{{3}}^{{\frac{{10}}{{7}}}}}$.

Now, it is time to exercise.

Exercise 1. Find ${{\left({{3}}^{{5}}\right)}}^{{2}}$.

Answer: ${{3}}^{{10}}$.

Exercise 2. Find ${{\left({{5}}^{{5}}\right)}}^{{-{2}}}$.

Answer: ${{5}}^{{-{10}}}=\frac{{1}}{{{5}}^{{10}}}$.

Exercise 3. Find ${{\left({{4}}^{{\frac{{3}}{{5}}}}\right)}}^{{5}}$.

Answer: ${{4}}^{{3}}={64}$.

Exercise 4. Find ${{\left({{3}}^{{2}}\right)}}^{{-\frac{{1}}{{5}}}}$.

Answer: ${{3}}^{{-\frac{{2}}{{5}}}}=\frac{{1}}{{\sqrt[{{5}}]{{\frac{{1}}{{9}}}}}}$.

Exercise 5. Find ${{\left({\sqrt[{{7}}]{{\frac{{1}}{{27}}}}}\right)}}^{{2}}$.

Answer: ${{\left({\sqrt[{{7}}]{{{{3}}^{{-{3}}}}}}\right)}}^{{2}}={{\left({{3}}^{{-\frac{{3}}{{7}}}}\right)}}^{{2}}={{3}}^{{-\frac{{6}}{{7}}}}=\frac{{1}}{{{3}}^{{\frac{{6}}{{7}}}}}$.