1.2 Exponents and scientific notation (Page 4/9)

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Try It

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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Questions & Answers

Why is b in the answer

Dahsolar Reply

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Brad Reply

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Nitin

(Pcos∅+qsin∅)/(pcos∅-psin∅)

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Chabelita

solve for X,,4^X-6(2^)-16=0

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x4xminus 2

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ZAHRO Reply

If  , , are the roots of the equation 3 2 0, x px qx r     Find the value of 1  .

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Parts of a pole were painted red, blue and yellow. 3/5 of the pole was red and 7/8 was painted blue. What part was painted yellow?

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Parts of the pole was painted red, blue and yellow. 3 /5 of the pole was red and 7 /8 was painted blue. What part was painted yellow?

Patrick

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Katleho Reply

Lairene and Mae are joking that their combined ages equal Sam’s age. If Lairene is twice Mae’s age and Sam is 69 yrs old, what are Lairene’s and Mae’s ages?

Mary Reply

23yrs

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christopher

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Alieu

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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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