# 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.

1. $\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}$
2. $\frac{{f}^{47}}{{f}^{49}\cdot f}$
3. $\frac{2{k}^{4}}{5{k}^{7}}$
1. $\frac{1}{{\left(-3t\right)}^{6}}$
2. $\frac{1}{{f}^{3}}$
3. $\frac{2}{5{k}^{3}}$

## 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.

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

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

1. ${t}^{-11}\cdot {t}^{6}$
2. $\frac{{25}^{12}}{{25}^{13}}$
1. ${t}^{-5}=\frac{1}{{t}^{5}}$
2. $\frac{1}{25}$

## 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 $\text{\hspace{0.17em}}{\left(pq\right)}^{3}.\text{\hspace{0.17em}}$ We begin by using the associative and commutative properties of multiplication to regroup the factors.

In other words, $\text{\hspace{0.17em}}{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}.$

## The power of a product rule of exponents

For any real numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and any integer $\text{\hspace{0.17em}}n,$ the power of a product rule of exponents states that

${\left(ab\right)}^{n}={a}^{n}{b}^{n}$

## 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.

1. ${\left(a{b}^{2}\right)}^{3}$
2. ${\left(2t\right)}^{15}$
3. ${\left(-2{w}^{3}\right)}^{3}$
4. $\frac{1}{{\left(-7z\right)}^{4}}$
5. ${\left({e}^{-2}{f}^{2}\right)}^{7}$

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

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

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

1. ${\left({g}^{2}{h}^{3}\right)}^{5}$
2. ${\left(5t\right)}^{3}$
3. ${\left(-3{y}^{5}\right)}^{3}$
4. $\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}$
5. ${\left({r}^{3}{s}^{-2}\right)}^{4}$
1. ${g}^{10}{h}^{15}$
2. $125{t}^{3}$
3. $-27{y}^{15}$
4. $\frac{1}{{a}^{18}{b}^{21}}$
5. $\frac{{r}^{12}}{{s}^{8}}$

## 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.

${\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}$

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

$\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$

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

$\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$

## The power of a quotient rule of exponents

For any real numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and any integer $\text{\hspace{0.17em}}n,$ the power of a quotient rule of exponents states that

${\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}$

## 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.

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

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