1.2 Exponents and scientific notation (Page 4/9)

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Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

$\frac{{(−3 t)}^{2}}{{(−3 t)}^{8}}$
$\frac{f^{47}}{f^{49} \cdot f}$
$\frac{2 k^{4}}{5 k^{7}}$

$\frac{1}{{(−3 t)}^{6}}$
$\frac{1}{f^{3}}$
$\frac{2}{5 k^{3}}$

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Using the product and quotient rules

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

$b^{2} \cdot b^{−8}$
${(- x)}^{5} \cdot {(- x)}^{−5}$
$\frac{−7 z}{{(−7 z)}^{5}}$

$b^{2} \cdot b^{−8} = b^{2 - 8} = b^{−6} = \frac{1}{b^{6}}$
${(- x)}^{5} \cdot {(- x)}^{−5} = {(- x)}^{5 - 5} = {(- x)}^{0} = 1$
$\frac{−7 z}{{(−7 z)}^{5}} = \frac{{(−7 z)}^{1}}{{(−7 z)}^{5}} = {(−7 z)}^{1 - 5} = {(−7 z)}^{−4} = \frac{1}{{(−7 z)}^{4}}$

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Try It

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

$t^{−11} \cdot t^{6}$
$\frac{25^{12}}{25^{13}}$

$t^{−5} = \frac{1}{t^{5}}$
$\frac{1}{25}$

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Finding the power of a product

To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider ${(p q)}^{3} .$ We begin by using the associative and commutative properties of multiplication to regroup the factors.

\begin{matrix} {(p q)}^{3} & = & \overset{3 factors}{\overset{}{(p q) \cdot (p q) \cdot (p q)}} \\ = & p \cdot q \cdot p \cdot q \cdot p \cdot q \\ = & \overset{3 factors}{\overset{}{p \cdot p \cdot p}} \cdot \overset{3 factors}{\overset{}{q \cdot q \cdot q}} \\ = & p^{3} \cdot q^{3} \end{matrix}

In other words, ${(p q)}^{3} = p^{3} \cdot q^{3} .$

A General Note

The power of a product rule of exponents

For any real numbers $a$ and $b$ and any integer $n,$ the power of a product rule of exponents states that

{(a b)}^{n} = a^{n} b^{n}

Using the power of a product rule

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

${(a b^{2})}^{3}$
${(2 t)}^{15}$
${(−2 w^{3})}^{3}$
$\frac{1}{{(−7 z)}^{4}}$
${(e^{−2} f^{2})}^{7}$

Use the product and quotient rules and the new definitions to simplify each expression.

${(a b^{2})}^{3} = {(a)}^{3} \cdot {(b^{2})}^{3} = a^{1 \cdot 3} \cdot b^{2 \cdot 3} = a^{3} b^{6}$
${(2 t)}^{15} = {(2)}^{15} \cdot {(t)}^{15} = 2^{15} t^{15} = 32, 768 t^{15}$
${(−2 w^{3})}^{3} = {(−2)}^{3} \cdot {(w^{3})}^{3} = −8 \cdot w^{3 \cdot 3} = −8 w^{9}$
$\frac{1}{{(−7 z)}^{4}} = \frac{1}{{(−7)}^{4} \cdot {(z)}^{4}} = \frac{1}{2, 401 z^{4}}$
${(e^{−2} f^{2})}^{7} = {(e^{−2})}^{7} \cdot {(f^{2})}^{7} = e^{−2 \cdot 7} \cdot f^{2 \cdot 7} = e^{−14} f^{14} = \frac{f^{14}}{e^{14}}$

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Try It

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

${(g^{2} h^{3})}^{5}$
${(5 t)}^{3}$
${(−3 y^{5})}^{3}$
$\frac{1}{{(a^{6} b^{7})}^{3}}$
${(r^{3} s^{−2})}^{4}$

$g^{10} h^{15}$
$125 t^{3}$
$−27 y^{15}$
$\frac{1}{a^{18} b^{21}}$
$\frac{r^{12}}{s^{8}}$

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Finding the power of a quotient

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.

{(e^{−2} f^{2})}^{7} = \frac{f^{14}}{e^{14}}

Let’s rewrite the original problem differently and look at the result.

\begin{matrix} {(e^{−2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{f^{14}}{e^{14}} \end{matrix}

It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

\begin{matrix} {(e^{- 2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{{(f^{2})}^{7}}{{(e^{2})}^{7}} \\ = & \frac{f^{2 \cdot 7}}{e^{2 \cdot 7}} \\ = & \frac{f^{14}}{e^{14}} \end{matrix}

A General Note

The power of a quotient rule of exponents

For any real numbers $a$ and $b$ and any integer $n,$ the power of a quotient rule of exponents states that

{(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}

Using the power of a quotient rule

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.

${(\frac{4}{z^{11}})}^{3}$
${(\frac{p}{q^{3}})}^{6}$
${(\frac{−1}{t^{2}})}^{27}$
${(j^{3} k^{−2})}^{4}$
${(m^{−2} n^{−2})}^{3}$

${(\frac{4}{z^{11}})}^{3} = \frac{{(4)}^{3}}{{(z^{11})}^{3}} = \frac{64}{z^{11 \cdot 3}} = \frac{64}{z^{33}}$
${(\frac{p}{q^{3}})}^{6} = \frac{{(p)}^{6}}{{(q^{3})}^{6}} = \frac{p^{1 \cdot 6}}{q^{3 \cdot 6}} = \frac{p^{6}}{q^{18}}$
${(\frac{−1}{t^{2}})}^{27} = \frac{{(−1)}^{27}}{{(t^{2})}^{27}} = \frac{−1}{t^{2 \cdot 27}} = \frac{−1}{t^{54}} = - \frac{1}{t^{54}}$
${(j^{3} k^{−2})}^{4} = {(\frac{j^{3}}{k^{2}})}^{4} = \frac{{(j^{3})}^{4}}{{(k^{2})}^{4}} = \frac{j^{3 \cdot 4}}{k^{2 \cdot 4}} = \frac{j^{12}}{k^{8}}$
${(m^{−2} n^{−2})}^{3} = {(\frac{1}{m^{2} n^{2}})}^{3} = \frac{{(1)}^{3}}{{(m^{2} n^{2})}^{3}} = \frac{1}{{(m^{2})}^{3} {(n^{2})}^{3}} = \frac{1}{m^{2 \cdot 3} \cdot n^{2 \cdot 3}} = \frac{1}{m^{6} n^{6}}$

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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