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