Using the Product Rule of Exponents

Consider the product ${x}^{3}\cdot {x}^{4}$ . Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.

\begin{matrix} x^{3} \cdot x^{4} & = \overset{3 factors 4 factors}{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x} \\ = \overset{7 factors}{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x} \\ = x^{7} \end{matrix}

$\begin{array}\text{ }x^{3}\cdot x^{4}\hfill&=\stackrel{\text{3 factors } \text{ 4 factors}}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x} \\ \hfill& =\stackrel{7 \text{ factors}}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x} \\ \hfill& =x^{7}\end{array}$

The result is that ${x}^{3}\cdot {x}^{4}={x}^{3+4}={x}^{7}$ .

Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents.

a^{m} \cdot a^{n} = a^{m + n}

${a}^{m}\cdot {a}^{n}={a}^{m+n}$

Now consider an example with real numbers.

2^{3} \cdot 2^{4} = 2^{3 + 4} = 2^{7}

${2}^{3}\cdot {2}^{4}={2}^{3+4}={2}^{7}$

We can always check that this is true by simplifying each exponential expression. We find that ${2}^{3}\\$ is 8, ${2}^{4}$ is 16, and ${2}^{7}$ is 128. The product $8\cdot 16$ equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.

A General Note: The Product Rule of Exponents

For any real number $a$ and natural numbers $m$ and $n$ , the product rule of exponents states that

a^{m} \cdot a^{n} = a^{m + n}

${a}^{m}\cdot {a}^{n}={a}^{m+n}$

Example 1: Using the Product Rule

Write each of the following products with a single base. Do not simplify further.

${t}^{5}\cdot {t}^{3}\\$
$\left(-3\right)^{5}\cdot \left(-3\right)$
${x}^{2}\cdot {x}^{5}\cdot {x}^{3}\\$

Solution

Use the product rule to simplify each expression.

${t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}$
${\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}$
${x}^{2}\cdot {x}^{5}\cdot {x}^{3}\\$

At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.

x^{2} \cdot x^{5} \cdot x^{3} = (x^{2} \cdot x^{5}) \cdot x^{3} = (x^{2 + 5}) \cdot x^{3} = x^{7} \cdot x^{3} = x^{7 + 3} = x^{10}

${x}^{2}\cdot {x}^{5}\cdot {x}^{3}=\left({x}^{2}\cdot {x}^{5}\right)\cdot {x}^{3}=\left({x}^{2+5}\right)\cdot {x}^{3}={x}^{7}\cdot {x}^{3}={x}^{7+3}={x}^{10}$

Notice we get the same result by adding the three exponents in one step.

x^{2} \cdot x^{5} \cdot x^{3} = x^{2 + 5 + 3} = x^{10}

${x}^{2}\cdot {x}^{5}\cdot {x}^{3}={x}^{2+5+3}={x}^{10}$

Try It 1

Write each of the following products with a single base. Do not simplify further.

a. ${k}^{6}\cdot {k}^{9}$
b. ${\left(\frac{2}{y}\right)}^{4}\cdot \left(\frac{2}{y}\right)$
c. ${t}^{3}\cdot {t}^{6}\cdot {t}^{5}$

Solution

Exponents and Scientific Notation