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Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
$\frac{{\left(\mathrm{-3}t\right)}^{2}}{{\left(\mathrm{-3}t\right)}^{8}}$
$\frac{{f}^{47}}{{f}^{49}\cdot f}$
$\frac{2{k}^{4}}{5{k}^{7}}$
$\frac{1}{{\left(\mathrm{-3}t\right)}^{6}}$
$\frac{1}{{f}^{3}}$
$\frac{2}{5{k}^{3}}$ Got questions? Get instant answers now!
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.
${b}^{2}\cdot {b}^{\mathrm{-8}}$
${\left(-x\right)}^{5}\cdot {\left(-x\right)}^{\mathrm{-5}}$
$\frac{\mathrm{-7}z}{{\left(\mathrm{-7}z\right)}^{5}}$
${b}^{2}\cdot {b}^{\mathrm{-8}}={b}^{2-8}={b}^{\mathrm{-6}}=\frac{1}{{b}^{6}}$
${\left(-x\right)}^{5}\cdot {\left(-x\right)}^{\mathrm{-5}}={\left(-x\right)}^{5-5}={\left(-x\right)}^{0}=1$
$\frac{\mathrm{-7}z}{{\left(\mathrm{-7}z\right)}^{5}}=\frac{{\left(\mathrm{-7}z\right)}^{1}}{{\left(\mathrm{-7}z\right)}^{5}}={\left(\mathrm{-7}z\right)}^{1-5}={\left(\mathrm{-7}z\right)}^{\mathrm{-4}}=\frac{1}{{\left(\mathrm{-7}z\right)}^{4}}$ Got questions? Get instant answers now! Got questions? Get instant answers now!
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Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
${t}^{\mathrm{-11}}\cdot {t}^{6}$
$\frac{{25}^{12}}{{25}^{13}}$
${t}^{\mathrm{-5}}=\frac{1}{{t}^{5}}$
$\frac{1}{25}$ Got questions? Get instant answers now!
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.
$$\begin{array}{ccc}\hfill {(pq)}^{3}& =& \stackrel{3\text{factors}}{\stackrel{}{(pq)\cdot (pq)\cdot (pq)}}\hfill \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\hfill \\ & =& \stackrel{3\text{factors}}{\stackrel{}{p\cdot p\cdot p}}\cdot \stackrel{3\text{factors}}{\stackrel{}{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}$$
In other words,
$\text{\hspace{0.17em}}{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}.$
A General Note
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.
${\left(a{b}^{2}\right)}^{3}$
${\left(2t\right)}^{15}$
${\left(\mathrm{-2}{w}^{3}\right)}^{3}$
$\frac{1}{{\left(\mathrm{-7}z\right)}^{4}}$
${\left({e}^{\mathrm{-2}}{f}^{2}\right)}^{7}$
Use the product and quotient rules and the new definitions to simplify each expression.
${\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}$
${\left(2t\right)}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}$
${\left(\mathrm{-2}{w}^{3}\right)}^{3}={\left(\mathrm{-2}\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=\mathrm{-8}\cdot {w}^{3\cdot 3}=\mathrm{-8}{w}^{9}$
$\frac{1}{{\left(\mathrm{-7}z\right)}^{4}}=\frac{1}{{\left(\mathrm{-7}\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}$
${\left({e}^{\mathrm{-2}}{f}^{2}\right)}^{7}={\left({e}^{\mathrm{-2}}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{\mathrm{-2}\cdot 7}\cdot {f}^{2\cdot 7}={e}^{\mathrm{-14}}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}$ Got questions? Get instant answers now! Got questions? Get instant answers now!
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Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
${\left({g}^{2}{h}^{3}\right)}^{5}$
${\left(5t\right)}^{3}$
${\left(\mathrm{-3}{y}^{5}\right)}^{3}$
$\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}$
${\left({r}^{3}{s}^{\mathrm{-2}}\right)}^{4}$
${g}^{10}{h}^{15}$
$125{t}^{3}$
$\mathrm{-27}{y}^{15}$
$\frac{1}{{a}^{18}{b}^{21}}$
$\frac{{r}^{12}}{{s}^{8}}$ Got questions? Get instant answers now!
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.
${\left({e}^{\mathrm{-2}}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}$
Let’s rewrite the original problem differently and look at the result.
$$\begin{array}{ccc}\hfill {\left({e}^{\mathrm{-2}}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$$
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
$$\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{({f}^{2})}^{7}}{{({e}^{2})}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$$
A General Note
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.
${\left(\frac{4}{{z}^{11}}\right)}^{3}$
${\left(\frac{p}{{q}^{3}}\right)}^{6}$
${\left(\frac{\mathrm{-1}}{{t}^{2}}\right)}^{27}$
${\left({j}^{3}{k}^{\mathrm{-2}}\right)}^{4}$
${\left({m}^{\mathrm{-2}}{n}^{\mathrm{-2}}\right)}^{3}$
${\left(\frac{4}{{z}^{11}}\right)}^{3}=\frac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\frac{64}{{z}^{11\cdot 3}}=\frac{64}{{z}^{33}}$
${\left(\frac{p}{{q}^{3}}\right)}^{6}=\frac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\frac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\frac{{p}^{6}}{{q}^{18}}$
${\left(\frac{\mathrm{-1}}{{t}^{2}}\right)}^{27}=\frac{{\left(\mathrm{-1}\right)}^{27}}{{\left({t}^{2}\right)}^{27}}=\frac{\mathrm{-1}}{{t}^{2\cdot 27}}=\frac{\mathrm{-1}}{{t}^{54}}=-\frac{1}{{t}^{54}}$
${\left({j}^{3}{k}^{\mathrm{-2}}\right)}^{4}={\left(\frac{{j}^{3}}{{k}^{2}}\right)}^{4}=\frac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\frac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\frac{{j}^{12}}{{k}^{8}}$
${\left({m}^{\mathrm{-2}}{n}^{\mathrm{-2}}\right)}^{3}={\left(\frac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\frac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\frac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\frac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\frac{1}{{m}^{6}{n}^{6}}$ Got questions? Get instant answers now! Got questions? Get instant answers now!