0.02: Review - whole number exponents

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Using the Product Rule of Exponents

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

\[ \begin{align*} x^3 \times x^4 &= \overbrace{x \times x \times x}^{\text{3 factors}} \times \overbrace{ x \times x \times x\times x}^{\text{4 factors}} \\[4pt] &= \overbrace{x\times x\times x\times x\times x\times x\times x}^{\text{7 factors}} \\[4pt] &=x^7 \end{align*}\]

Notice that the exponent of a product is the sum of the exponents of the two factors. 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.

\[ x^3\times x^4=x^{3+4}=x^7 \nonumber\]

Now consider an example with real numbers.

$2^3\times2^4=2^{3+4}=2^7$

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 $8\times16$ 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.

THE PRODUCT RULE OF EXPONENTS

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

\[a^m\times a^n=a^{m+n} \label{prod}\]

Example $\PageIndex{1}$: Using the Product Rule

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

$t^5\times t^3$
$(−3)^5\times(−3)$
$x^2\times x^5\times x^3$

Solution

Use the product rule (Equation \ref{prod}) to simplify each expression.

$t^5\times t^3=t^{5+3}=t^8$
$(−3)^5\times(−3)=(−3)^5\times(−3)^1=(−3)^{5+1}=(−3)^6$
$x^2\times x^5\times x^3$

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.

\[x^2\times x^5\times x^3=(x^2\times x^5) \times x^3=(x^{2+5})\times x^3=x^7\times x^3=x^{7+3}=x^{10} \nonumber\]

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

\[x^2\times x^5\times x^3=x^{2+5+3}=x^{10} \nonumber\]

$try-it.png$ Try It $\PageIndex{1}$

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

$k^6\times k^9$
$\left(\dfrac{2}{y}\right)^4\times\left(\dfrac{2}{y}\right)$
$t^3\times t^6\times t^5$

Answers: a. $k^{15}$ $ \qquad $ b. $\left(\dfrac{2}{y}\right)^5$ $ \qquad $ c. $t^{14}$

Using the Power Rule of Exponents

Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression $(x^2)^3$ . The expression inside the parentheses is multiplied twice because it has an exponent of $2$. Then the result is multiplied three times because the entire expression has an exponent of $3$.

\[\begin{align*} (x^2)^3 &= (x^2)\times(x^2)\times(x^2)\\ &= x\times x\times x\times x\times x\times x\\ &= x^6 \end{align*}\]

The exponent of the answer is the product of the exponents. In other words, when raising an exponential expression to a power, we write the result with the common base and multiply the exponents.

\[ (x^2)^3=x^{2⋅3}=x^6 \nonumber\]

Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.

Product Rule	Power Rule
$5^3\times5^4=5^{3+4}=5^7$	$(5^3)^4=5^{3\times4}=5^{12}$
$x^5\times x^2=x^{5+2}=x^7$	$(x^5)^2=x^{5\times2}=x^{10}$
$(3a)^7\times(3a)^{10}=(3a)^{7+10}=(3a)^{17}$	$((3a)^7)^{10}=(3a)^{7\times10}=(3a)^{70}$

THE POWER RULE OF EXPONENTS

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

\[(a^m)^n=a^{m⋅n} \label{power}\]

Example $\PageIndex{2}$: Using the Power Rule

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

$(x^2)^7$
$((2t)^5)^3$
$((−3)^5)^{11}$

Solution

Use the power rule (Equation \ref{power}) to simplify each expression.

$(x^2)^7=x^{2⋅7}=x^{14}$
$((2t)^5)^3=(2t)^{5⋅3}=(2t)^{15}$
$((−3)^5)^{11}=(−3)^{5⋅11}=(−3)^{55}$

$try-it.png$ Try It $\PageIndex{2}$

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

$ ((3y)^8)^3 $
$ (t^5)^7 $
$ ((-g)^4)^4 $

Answers: a. $ (3y)^{24} $ $ \qquad $ b. $ t^5 $ $ \qquad $ c. $ (-g)^{16} $

Using the Quotient Rule of Exponents

The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as $\dfrac{y^m}{y^n}$. Consider the example $\dfrac{y^9}{y^5}$ . Perform the division by canceling common factors.

\[\begin{align*} \dfrac{y^9}{y^5} &= \dfrac{y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y\cdot y\cdot y}\\ &= \dfrac{y\cdot y\cdot y\cdot y}{1}\\ &= y^4 \end{align*}\]

Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend. In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.

$\dfrac{y^9}{y^5}=y^{9−5}=y^4$

THE QUOTIENT RULE OF EXPONENTS

For any real number $a, a \ne 0$ and natural numbers $m$ and $n$, the quotient rule of exponents states that

\[\dfrac{a^m}{a^n}=a^{m−n} \label{quot}\]

Instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.

Example $\PageIndex{3}$: Using the Quotient Rule

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

$\dfrac{(−2)^{14}}{(−2)^{9}}$
$\dfrac{t^{23}}{t^{15}}$
$\dfrac{(z\sqrt{2})^5}{z\sqrt{2}}$

Solution

Use the quotient rule (Equation \ref{quot}) to simplify each expression.

$\dfrac{(−2)^{14}}{(−2)^{9}}=(−2)^{14−9}=(−2)^5$
$\dfrac{t^{23}}{t^{15}}=t^{23−15}=t^8$
$\dfrac{(z\sqrt{2})^5}{z\sqrt{2}}=(z\sqrt{2})^{5−1}=(z\sqrt{2})^4$

$try-it.png$ Try It $\PageIndex{3}$

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

$\dfrac{s^{75}}{s^{68}}$
$\dfrac{(−3)^6}{−3}$
$\dfrac{(ef^2)^5}{(ef^2)^3}$

Answers: a. $s^7$ $ \qquad $ b. $(−3)^5$ $ \qquad $ c. $(ef^2)^2$

Using the Zero Exponent Rule of Exponents

What would happen if the Quotient Rule were used and $m=n$? Consider the example.

\[\dfrac{t^8}{t^8}=1 \qquad \text{because a number divided by itself is 1}\nonumber\]

If we were to simplify the expression using the quotient rule instead, we would have

\[\dfrac{t^8}{t^8}=t^{8−8}=t^0 \nonumber\]

If we equate the two answers, the result is $t^0=1$. This is true for any nonzero real number, or any variable representing a real number. The sole exception is the expression $0^0$, whose value is undefined.

THE ZERO EXPONENT RULE OF EXPONENTS

For any nonzero real number $a$, the zero exponent rule of exponents states that

\[a^0=1\]

Example $\PageIndex{4}$: Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.

$\dfrac{c^3}{c^3}$
$\dfrac{-3x^5}{x^5}$
$\dfrac{(j^2k)^4}{(j^2k)\times(j^2k)^3}$
$\dfrac{5(rs^2)^2}{(rs^2)^2}$

Solution

Use the zero exponent and other rules to simplify each expression.

a. \[\begin{align*} \dfrac{c^3}{c^3} &= c^{3-3}\\ &= c^0\\ &= 1 \end{align*}\]

b. \[\begin{align*} \dfrac{-3x^5}{x^5} &= -3\times\dfrac{x^5}{x^5}\\ &= -3\times x^{5-5}\\ &= -3\times x^0\\ &= -3\times 1\\ &= -3 \end{align*}\]

c. \[\begin{align*} \dfrac{(j^2k)^4}{(j^2k)\times(j^2k)^3} &= \dfrac{(j^2k)^4}{(j^2k)^{1+3}} && \text{ Use the product rule in the denominator}\\ &= \dfrac{(j^2k)^4}{(j^2k)^4} && \text{ Simplify}\\ &= (j^2k)^{4-4} && \text{ Use the quotient rule}\\ &= (j^2k)^0 && \text{ Simplify}\\ &= 1 \end{align*}\]

d. \[\begin{align*} \dfrac{5(rs^2)^2}{(rs^2)^2} &= 5(rs^2)^{2-2} && \text{ Use the quotient rule}\\ &= 5(rs^2)^0 && \text{ Simplify}\\ &= 5\times1 && \text{ Use the zero exponent rule}\\ &= 5 && \text{ Simplify} \end{align*}\]

$try-it.png$ Try It $\PageIndex{4}$

Simplify each expression using the zero exponent rule of exponents.

$\dfrac{t^7}{t^7}$
$\dfrac{(de^2)^{11}}{2(de^2)^{11}}$
$\dfrac{w^4\times w^2}{w^6}$
$\dfrac{t^3\times t^4}{t^2\times t^5}$

Answers: a. $1$ $ \qquad $ b. $\dfrac{1}{2}$ $ \qquad $ c. $1$ $ \qquad $ d. $1$

Using the Negative Rule of Exponents

Consider the situation where one exponential expression is divided by another exponential expression with a larger exponent. For example, $\dfrac{t^3}{t^5}$.

\[\begin{align*} \dfrac{t^3}{t^5} &= \dfrac{t\times t\times t}{t\times t\times t\times t\times t} \\ &= \dfrac{1}{t\times t}\\ &= \dfrac{1}{t^2} \end{align*}\]

If we were to simplify the original expression using the quotient rule, we would have

\[\begin{align*} \dfrac{t^3}{t^5} &= t^{3-5} \\ &= t^{-2} \end{align*}\]

Putting the answers together, we have $t^{−2}=\dfrac{1}{t^2}$. This is true for any nonzero real number $t$.

In general, a factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.

$a^{−n}=\dfrac{1}{a^n}$ and $a^n=\dfrac{1}{a^{−n}}$

We have shown that the exponential expression $a^n$ is defined when $n$ is a natural number, $0$, or the negative of a natural number. That means that $a^n$ is defined for any integer $n$. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n$.

THE NEGATIVE RULE OF EXPONENTS

For any nonzero real number $a$ and integer $n$, the negative rule of exponents states that

\[a^{−n}=\dfrac{1}{a^n} \qquad \qquad \text{ and } \qquad \qquad a^n=\dfrac{1}{a^{−n}} \]

Example $\PageIndex{5}$: Using the Negative Exponent Rule

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

$\dfrac{\theta^3}{\theta^{10}}$
$\dfrac{z^2\times z}{z^4}$
$\dfrac{(-5t^3)^4}{(-5t^3)^8}$

Solution

$\dfrac{\theta^3}{\theta^{10}}=\theta^{3-10}=\theta^{-7}=\dfrac{1}{\theta^7}$
$\dfrac{z^2\times z}{z^4}=\dfrac{z^{2+1}}{z^4}=\dfrac{z^3}{z^4}=z^{3-4}=z^{-1}=\dfrac{1}{z}$
$\dfrac{(-5t^3)^4}{(-5t^3)^8}=(-5t^3)^{4-8}=(-5t^3)^{-4}=\dfrac{1}{(-5t^3)^4}$

$try-it.png$ Try It $\PageIndex{5}$

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

$\dfrac{(-3t)^2}{(-3t)^8}$
$\dfrac{f^{47}}{f^{49}\times f}$
$\dfrac{2k^4}{5k^7}$

Answers: a. $\dfrac{1}{(-3t)^6}$ $ \qquad $ b. $\dfrac{1}{f^3}$ $ \qquad $ c. $\dfrac{2}{5k^3}$

Example $\PageIndex{6}$: 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.

$b^2\times b^{-8}$
$(-x)^5\times(-x)^{-5}$
$\dfrac{-7z}{(-7z)^5}$

Solution

$b^2\times b^{-8}=b^{2\: + \: -8}=b^{-6}=\dfrac{1}{b^6}$
$(-x)^5\times(-x)^{-5}=(-x)^{5\: + \:-5}=(-x)^0=1$
$\dfrac{-7z}{(-7z)^5}= \dfrac{(-7z)^1}{(-7z)^5}=(-7z)^{1\:-\:5}=(-7z)^{-4}=\dfrac{1}{(-7z)^4}$

$try-it.png$ Try It $\PageIndex{6}$

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

$t^{-11}\times t^6$
$\dfrac{25^{12}}{25^{13}}$

Answers: a. $t^{-5}=\dfrac{1}{t^5}$ $ \qquad $ b. $\dfrac{1}{25}$

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 $(pq)^3$. We begin by using the associative and commutative properties of multiplication to regroup the factors.

\[\begin{align*} (pq)^3 &= (pq)\times(pq)\times(pq)\\ &= p\times q\times p\times q\times p\times q\\ &= p^3\times q^3 \end{align*}\]

In other words, $(pq)^3=p^3\times q^3$.

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

\[(ab)^n=a^nb^n\]

Example $\PageIndex{7}$: 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.

$(ab^2)^3$
$(2t)^{15}$
$(-2w^3)^3$
$\dfrac{1}{(-7z)^4}$
$(e^{-2}f^2)^7$

Solution

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

a. $(ab^2)^3=(a)^3\times(b^2)^3=a^{1\times3}\times b^{2\times3}=a^3b^6$

b. $(2t)^{15}=(2)^{15}\times(t)^{15}=2^{15}t^{15}=32,768t^{15}$

c. $(−2w^3)^3=(−2)^3\times(w^3)^3=−8\times w^{3\times3}=−8w^9$

d. $\dfrac{1}{(-7z)^4}=\dfrac{1}{(-7)^4\times(z)^4}=\dfrac{1}{2401z^4}$

e. $(e^{-2}f^2)^7=(e^{−2})^7\times(f^2)^7=e^{−2\times7}\times f^{2\times7}=e^{−14}f^{14}=\dfrac{f^{14}}{e^{14}}$

$try-it.png$ Try It $\PageIndex{7}$

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

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

Answers: a. $g^{10}h^{15}$ $ \qquad $ b. $125t^3$ $ \qquad $ c. $-27y^{15}$ $ \qquad $ d. $\dfrac{1}{a^{18}b^{21}}$ $ \qquad $ e. $\dfrac{r^{12}}{s^8}$

Finding the Power of a Quotient

To simplify the power of a quotient of two expressions, consider the expression below.

$ \left( \dfrac{2}{x} \right)^3 = \dfrac{2}{x} \cdot \dfrac{2}{x} \cdot \dfrac{2}{x} = \dfrac{2 \cdot 2 \cdot 2}{x \cdot x \cdot x} = \dfrac{2^3}{x^3} $

Thus in general, the power of a quotient of factors is the quotient of the powers of the factors.

THE POWER OF A QUOTIENT RULE OF EXPONENTS

For any real numbers $a$ and $b$, $b \ne 0 $, and any integer $n$, the power of a quotient rule of exponents states that

\[\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\]

Example $\PageIndex{8}$: 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.

$\left(\dfrac{4}{z^{11}}\right)^3$
$\left(\dfrac{p}{q^3}\right)^6$
$\left(\dfrac{-1}{t^2}\right)^{27}$
$(j^3k^{-2})^4$
$(m^{-2}n^{-2})^3$

Solution

a. $\left(\dfrac{4}{z^{11}}\right)^3=\dfrac{(4)^3}{(z^{11})^3}=\dfrac{64}{z^{11\times3}}=\dfrac{64}{z^{33}}$

b. $\left(\dfrac{p}{q^3}\right)^6=\dfrac{(p)^6}{(q^3)^6}=\dfrac{p^{1\times6}}{q^{3\times6}}=\dfrac{p^6}{q^{18}}$

c. $\left(\dfrac{-1}{t^2}\right)^{27}=\dfrac{(-1)^{27}}{(t^2)^{27}}=\dfrac{-1}{t^{2\times27}}=\dfrac{-1}{t^{54}}=-\dfrac{1}{t^{54}}$

d. $(j^3k^{-2})^4=\left(\dfrac{j^3}{k^2}\right)^4=\dfrac{(j^3)^4}{(k^2)^4}=\dfrac{j^{3\times4}}{k^{2\times4}}=\dfrac{j^{12}}{k^8}$

e. $(m^{-2}n^{-2})^3=\left(\dfrac{1}{m^2n^2}\right)^3=\dfrac{(1)^3}{(m^2n^2)^3}=\dfrac{1}{(m^2)^3(n^2)^3}=\dfrac{1}{m^{2\times3}n^{2\times3}}=\dfrac{1}{m^6n^6}$

$try-it.png$ Try It $\PageIndex{8}$

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

$\left(\dfrac{b^5}{c}\right)^3$
$\left(\dfrac{5}{u^8}\right)^4$
$\left(\dfrac{-1}{w^3}\right)^{35}$
$(p^{-4}q^3)^8$
$(c^{-5}d^{-3})^4$

Answers: a. $\dfrac{b^{15}}{c^3}$ $ \qquad $ b. $\dfrac{625}{u^{32}}$ $ \qquad $ c. $\dfrac{-1}{w^{105}}$ $ \qquad $ d. $\dfrac{q^{24}}{p^{32}}$ $ \qquad $ e. $\dfrac{1}{c^{20}d^{12}}$

Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.

Example $\PageIndex{9}$: Simplifying Exponential Expressions

Simplify each expression and write the answer with positive exponents only.

$(6m^2n^{-1})^3 \\[4pt] $
$17^5\times17^{-4}\times17^{-3} \\[4pt] $
$\left(\dfrac{u^{-1}v}{v^{-1}}\right)^2 \\[4pt] $
$(-2a^3b^{-1})(5a^{-2}b^2) \\[4pt] $
$(x^2\sqrt{2})^4(x^2\sqrt{2})^{-4} \\[4pt] $
$\dfrac{(3w^2)^5}{(6w^{-2})^2}$

Solution

a. \[\begin{align*} (6m^2n^{-1})^3 &= (6)^3(m^2)^3(n^{-1})^3 && \text{ The power of a product rule}\\ &= 6^3m^{2\times3}n^{-1\times3} && \text{ The power rule}\\ &= 216m^6n^{-3} && \text{ The power rule}\\ &= \dfrac{216m^6}{n^3} && \text{ The negative exponent rule} \end{align*}\]

b. \[\begin{align*} 17^5\times17^{-4}\times17^{-3} &= 17^{5+(-4)+(-3)} && \text{ The product rule}\\ &= 17^{-2} && \text{ Simplify}\\ &= \dfrac{1}{17^2} \text{ or } \dfrac{1}{289} && \text{ The negative exponent rule} \end{align*}\]

To avoid making mistakes in subtracting negative numbers, it is easier to apply the Negative Exponent Rule before the Quotient Rule. Both approaches are illustrated below.

c. \[\begin{align*} \left ( \dfrac{u^{-1}v}{v^{-1}} \right )^2
&= \dfrac{(u^{-1}v)^2}{(v^{-1})^2} && \text{ The power of a quotient rule}\\
&= \dfrac{u^{-2}v^2}{v^{-2}} && \text{ The power of a product rule}\\
\end{align*}\]

$ \begin{array}{ll|ll}
= u^{-2}v^{2-(-2)} & \text{Quotient rule} & =\dfrac{v^2 {\color{Cerulean}{v^2}}}{ \color{Cerulean}{u^2} } & \text{Negative Exponent Rule}\\
= u^{-2}v^4 & \text{Simplify} & =\dfrac{v^{2+2}}{u^2} & \text{Product Rule}\\
= \dfrac{v^4}{u^2} & \text{Negative exponent rule} & =\dfrac{v^4}{u^2} & \text{Simplify}\\
\end{array} $

d. \[\begin{align*} \left (-2a^3b^{-1} \right ) \left(5a^{-2}b^2 \right )
&= -2 \cdot a^3 \cdot b^{-1} \cdot 5 \cdot a^{-2} \cdot b^2 && \text{ Associative law of multiplication}\\
&= -2 \cdot 5 \cdot a^3 \cdot a^{-2} \cdot b^{-1} \cdot b^2 && \text{ Commutative law of multiplication}\\
&= (-2 \cdot 5) \cdot (a^3 \cdot a^{-2}) \cdot (b^{-1} \cdot b^2) && \text{ Associative law of multiplication}\\
&= -10\times a^{3+(-2)}\times b^{-1+2} && \text{ The product rule}\\
&= -10ab && \text{ Simplify} \end{align*}\]

e. \[\begin{align*} \left (x^2\sqrt{2})^4(x^2\sqrt{2} \right )^{-4} &= \left (x^2\sqrt{2} \right )^{4-4} && \text{ The product rule}\\ &= \left (x^2\sqrt{2} \right )^0 && \text{ Simplify}\\ &= 1 && \text{ The zero exponent rule} \end{align*}\]

f. \[\begin{align*} \dfrac{(3w^2)^5}{(6w^{-2})^2}
&= \dfrac{(3)^5\times(w^2)^5}{(6)^2\times(w^{-2})^2} && \text{ The power of a product rule}\\
&= \dfrac{3^5w^{2\times5}}{6^2w^{-2\times2}} && \text{ The power rule}\\
&= \dfrac{243w^{10}}{36w^{-4}} && \text{ Simplify}\\
&= \dfrac{243w^{10}w^4}{36} && \text{Negative exponent rule}\\
&= \dfrac{243w^{10+4}}{36} && \text{ The product rule}\\
&= \dfrac{27w^{14}}{4} && \text{ Reduce fraction} \end{align*}\]

$try-it.png$ Try It $\PageIndex{9x}$

Simplify each of the following exponential expressions. Write answers with positive exponents.

$ (2uv^{-2})^{-3} $
$ x^8 \cdot c^{-12} \cdot x $
$ \Big( \dfrac{e^2f^{-3}}{f^{-1}} \Big)^2 $
$ (9r^{-5}s^3)(3r^6s^{-4}) $
$ ( \frac{4}{9}tw^{-2} )^{-3} ( \frac{4}{9}tw^{-2} )^3 $
$ \dfrac{ (2h^2k)^4 }{ (7h^{-1}k^2)^2 } $

Answers: a. $\dfrac{v^6}{8u^3}$ $ \qquad $ b. $\dfrac{x^9}{c^12}$ $ \qquad $ c. $\dfrac{e^4}{f^4}$ $ \qquad $ d. $\dfrac{27r}{s}$ $ \qquad $ e. $ 1 $ $ \qquad $ f. $ \dfrac{8h^{16}}{49}$

Working "Backwards"

All our previous examples had a constant as an exponent and a variable in the base, like $x^2$. Expressions like this are called power functions. In contrast, the base can be a constant and the exponent can have a variable in it, like $2^x$. Expressions in this form are called exponential functions. When simplifying expressions with the variable in the exponent, we often use the Laws of Exponents "backwards". Some examples below illustrate this.

Example $\PageIndex{10}$

Write the following exponential expressions so they use only one exponent, and that exponent is just $x$.

1. $ 4^{x+2} $

2. $ 5^{x-3} $

3. $ 6^x \cdot 7^x $

4. $ \dfrac{9^x}{8^x} $

5. $ 4^{2x} $

Solution

Using the Product Rule in reverse we obtain $ 4^{x+2} = 4^x \cdot 4^2 = 16 \cdot 4^x $.
Using the Product Rule in reverse we obtain $ 5^{x-3} = 5^x \cdot 5^{-3} = \dfrac{5^x}{5^3} = \dfrac {5^x}{125} $
Using the Power Rule in reverse we obtain $ 6^x \cdot 7^x = ( 6 \times 7)^x = 42^x$
Using the Power Rule in reverse we obtain $ \dfrac{9^x}{8^x} = \left( \dfrac{9}{8} \right) ^x = 1.125^x $
Using the Power Rule in reverse we obtain $ 4^{2x} = (4^2)^x = 16^x $. Notice the difference between this expression and $ 4^{x+2} $!

In (3) and (4) we are using the 'like exponents rule" - if you have like exponents, then you multiply the bases and keep the exponent the same. Somewhat conversely, the Product Rule could be called the "like bases rule" - if you have like bases, then you add the exponents and keep the base the same.

Key Equations

Rules of Exponents For nonzero real numbers a and b and integers m and n
Product rule	$a^m⋅a^n=a^{m+n}$
Quotient rule	$\dfrac{a^m}{a^n}=a^{m−n}$
Power rule	$(a^m)^n=a^{m⋅n}$
Zero exponent rule	$a^0=1$
Negative rule	$a^{−n}=\dfrac{1}{a^n}$
Power of a product rule	$(a⋅b)^n=a^n⋅b^n$
Power of a quotient rule	$\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

Key Concepts

Products of exponential expressions with the same base can be simplified by adding exponents.
Quotients of exponential expressions with the same base can be simplified by subtracting exponents.
Powers of exponential expressions with the same base can be simplified by multiplying exponents.
An expression with exponent zero is defined as 1.
An expression with a negative exponent is defined as a reciprocal.
The power of a product of factors is the same as the product of the powers of the same factors.
The power of a quotient of factors is the same as the quotient of the powers of the same factors.
The rules for exponential expressions can be combined to simplify more complicated expressions.

Contributors

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.

Product Rule	Power Rule
\(5^3\times5^4=5^{3+4}=5^7\)	\((5^3)^4=5^{3\times4}=5^{12}\)
\(x^5\times x^2=x^{5+2}=x^7\)	\((x^5)^2=x^{5\times2}=x^{10}\)
\((3a)^7\times(3a)^{10}=(3a)^{7+10}=(3a)^{17}\)	\(((3a)^7)^{10}=(3a)^{7\times10}=(3a)^{70}\)

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Working "Backwards"

Key Equations

Key Concepts

Contributors

Rules of Exponents For nonzero real numbers a and b and integers m and n
Product rule	\(a^m⋅a^n=a^{m+n}\)
Quotient rule	\(\dfrac{a^m}{a^n}=a^{m−n}\)
Power rule	\((a^m)^n=a^{m⋅n}\)
Zero exponent rule	\(a^0=1\)
Negative rule	\(a^{−n}=\dfrac{1}{a^n}\)
Power of a product rule	\((a⋅b)^n=a^n⋅b^n\)
Power of a quotient rule	\(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\)