1.2: Exponents
- The Rules of Exponents
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.
\[ \begin{align*} x^3 \times x^4 &= \overbrace{x \times x \times x}^{\text{3 factors}} \times \overbrace{ x \times x \times x\times x}^{\text{4 factors}} \\[4pt] &= \overbrace{x\times x\times x\times x\times x\times x\times x}^{\text{7 factors}} \\[4pt] &=x^7 \end{align*}\]
The result is that \(x^3\times x^4=x^{3+4}=x^7\).
Notice that the exponent of the product is the sum of the exponents of the terms. 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.
\[a^m\times a^n=a^{m+n}\]
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.
For any real number a and natural numbers \(m\) and \(n\), the product rule of exponents states that
\[a^m\times a^n=a^{m+n} \label{prod}\]
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.
\[x^2\times x^5\times x^3=x^{2+5+3}=x^{10} \nonumber\]
Write each of the following products with a single base. Do not simplify further.
- \(k^6\times k^9\)
- \(\left(\dfrac{2}{y}\right)^4\times\left(\dfrac{2}{y}\right)\)
- \(t^3\times t^6\times t^5\)
- Answer a
-
\(k^{15}\)
- Answer b
-
\(\left(\dfrac{2}{y}\right)^5\)
- Answer c
-
\(t^{14}\)
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 \(\dfrac{y^m}{y^n}\), where \(m>n\). 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.
\[\dfrac{a^m}{a^n}=a^{m−n}\]
In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.
\(\dfrac{y^9}{y^5}=y^{9−5}=y^4\)
For the time being, we must be aware of the condition \(m>n\). Otherwise, the difference \(m-n\) could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.
For any real number \(a\) and natural numbers \(m\) and \(n\), such that \(m>n\), the quotient rule of exponents states that
\[\dfrac{a^m}{a^n}=a^{m−n} \label{quot}\]
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\)
- \(\dfrac{(z\sqrt{2})^5}{z\sqrt{2}}=(z\sqrt{2})^{5−1}=(z\sqrt{2})^4\)
Write each of the following products with a single base. Do not simplify further.
- \(\dfrac{s^{75}}{s^{68}}\)
- \(\dfrac{(−3)^6}{−3}\)
- \(\dfrac{(pq^2)^5}{(pq^2)^3}\)
- Answer a
-
\(s^7\)
- Answer b
-
\((−3)^5\)
- Answer c
-
\((pq^2)^2\)
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: \((x^2)^3=x^{2⋅3}=x^6\). In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.
\[(a^m)^n=a^{m⋅n}\]
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 |
|---|---|
| \(5^3\times5^4=5^{3+4}=5^7\) | \((5^3)^4=5^{3\times4}=5^{12}\) |
| \(x^5\times x^2=x^{5+2}=x^7\) | \((x^5)^2=x^{5\times2}=x^{10}\) |
| \((3a)^7\times(3a)^{10}=(3a)^{7+10}=(3a)^{17}\) | \(((3a)^7)^{10}=(3a)^{7\times10}=(3a)^{70}\) |
For any real number a and positive integers m and n, the power rule of exponents states that
\[(a^m)^n=a^{m⋅n} \label{power}\]
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.
- \((x^2)^7=x^{2⋅7}=x^{14}\)
- \(((2t)^5)^3=(2t)^{5⋅3}=(2t)^{15}\)
- \(((−3)^5)^{11}=(−3)^{5⋅11}=(−3)^{55}\)
Write each of the following products with a single base. Do not simplify further.
- \(((3y)^8)^3\)
- \((t^5)^7\)
- \(((−x)^4)^4\)
- Answer a
-
\((3y)^{24}\)
- Answer b
-
\(t^{35}\)
- Answer c
-
\((−x)^{16}\)
Using the Zero Exponent Rule of Exponents
Return to the quotient rule. We made the condition that \(m>n\) so that the difference \(m−n\) would never be zero or negative. What would happen if \(m=n\) ? In this case, we would use the zero exponent rule of exponents to simplify the expression to \(1\). To see how this is done, let us begin with an example.
\[\dfrac{t^8}{t^8}=1 \nonumber\]
If we were to simplify the original expression using the quotient rule, 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.
\[a^0=1 \nonumber\]
The sole exception is the expression \(0^0\). This appears later in more advanced courses, but for now, we will consider the value to be undefined.
For any nonzero real number a, the zero exponent rule of exponents states that
\[a^0=1\]
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. \[\begin{align*} \dfrac{5(rs^2)^2}{(rs^2)^2} &= 5(rs^2)^{2-2} && \text{ Use the quotient rule}\\ &= 5(rs^2)^0 && \text{ Simplify}\\ &= 5\times1 && \text{ Use the zero exponent rule}\\ &= 5 && \text{ Simplify} \end{align*}\]
Simplify each expression using the zero exponent rule of exponents.
- \(\dfrac{t^7}{t^7}\)
- \(\dfrac{(xy^2)^{11}}{2(xy^2)^{11}}\)
- \(\dfrac{w^4\times w^2}{w^6}\)
- \(\dfrac{t^3\times t^4}{t^2\times t^5}\)
- Answer a
-
\(1\)
- Answer b
-
\(\dfrac{1}{2}\)
- Answer c
-
\(1\)
- Answer d
-
\(1\)
Using the Negative Rule of Exponents
Another useful result occurs if we relax the condition that \(m>n\) in the quotient rule even further. For example, can we simplify \(\dfrac{t^3}{t^5}\)? When \(m<n\) —that is, where the difference \(m−n\) is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal.
Divide one exponential expression by another with a larger exponent. Use our example, \(\dfrac{t^3}{t^5}\).
\[\begin{align*} \dfrac{t^3}{t^5} &= \dfrac{t\times t\times t}{t\times t\times t\times t\times t} \\ &= \dfrac{1}{t\times t}\\ &= \dfrac{1}{h^2} \end{align*}\]
If we were to simplify the original expression using the quotient rule, we would have
\[\begin{align*} \dfrac{t^3}{t^5} &= h^{3-5} \\ &= h^{-2} \end{align*}\]
Putting the answers together, we have \(h^{−2}=\dfrac{1}{h^2}\). This is true for any nonzero real number, or any variable representing a nonzero real number.
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.
We have shown that the exponential expression an is defined when \(n\) is a natural number, \(0\), or the negative of a natural number. That means that an 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 \(n\).
For any nonzero real number a and natural number n, the negative rule of exponents states that
\[a^{−n}=\dfrac{1}{a^n}\]
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
- \(\dfrac{x^3}{x^{10}}\)
- \(\dfrac{z^2\times z}{z^4}\)
- \(\dfrac{(-5t^3)^4}{(-5t^3)^8}\)
Solution
- \(\dfrac{x^3}{x^{10}}=x^{3-10}=x^{-7}=\dfrac{1}{x^7}\)
- \(\dfrac{z^2\times z}{z^4}=\dfrac{z^{2+1}}{z^4}=\dfrac{z^3}{z^4}=z^{3-4}=z^{-1}=\dfrac{1}{z}\)
- \(\dfrac{(-5t^3)^4}{(-5t^3)^8}=(-5t^3)^{4-8}=(-5t^3)^{-4}=\dfrac{1}{(-5t^3)^4}\)
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
- \(\dfrac{(-3t)^2}{(-3t)^8}\)
- \(\dfrac{p^{47}}{p^{49}\times p}\)
- \(\dfrac{2k^4}{5k^7}\)
- Answer a
-
\(\dfrac{1}{(-3t)^6}\)
- Answer b
-
\(\dfrac{1}{p^3}\)
- Answer c
-
\(\dfrac{2}{5k^3}\)
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
- \(b^2\times b^{-8}=b^{2-8}=b^{-6}=\dfrac{1}{b^6}\)
- \((-x)^5\times(-x)^{-5}=(-x)^{5-5}=(-x)^0=1\)
- \(\dfrac{-7z}{(-7z)^5}= \dfrac{(-7z)^1}{(-7z)^5}=(-7z)^{1-5}=(-7z)^{-4}=\dfrac{1}{(-7z)^4}\)
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
- \(t^{-11}\times t^6\)
- \(\dfrac{25^{12}}{25^{13}}\)
- Answer a
-
\(t^{-5}=\dfrac{1}{t^5}\)
- Answer 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, \((pq)^3=p^3\times q^3\).
For any real numbers a and b and any integer n, the power of a product rule of exponents states that
\[(ab)^n=a^nb^n\]
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
- \((ab^2)^3\)
- \((2t)^{15}\)
- \((-2w^3)^3\)
- \(\dfrac{1}{(-7z)^4}\)
- \((p^{-2}q^2)^7\)
Solution
Use the product and quotient rules and the new definitions to simplify each expression.
- \((ab^2)^3=(a)^3\times(b^2)^3=a^{1\times3}\times b^{2\times3}=a^3b^6\)
- \((2t)^{15}=(2)^{15}\times(t)^{15}=2^{15}t^{15}=32,768t^{15}\)
- \((−2w^3)^3=(−2)^3\times(w^3)^3=−8\times w^{3\times3}=−8w^9\)
- \(\dfrac{1}{(-7z)^4}=\dfrac{1}{(-7)^4\times(z)^4}=\dfrac{1}{2401z^4}\)
- \((p^{-2}q^2)^7=(p^{−2})^7\times(q^2)^7=p^{−2\times7}\times q^{2\times7}=p^{−14}q^{14}=\dfrac{q^{14}}{p^{14}}\)
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
- \((x^2y^3)^5\)
- \((5t)^3\)
- \((-3y^5)^3\)
- \(\dfrac{1}{(a^6b^7)^3}\)
- \((r^3s^{-2})^4\)
- Answer a
-
\(x^{10}y^{15}\)
- Answer b
-
\(125t^3\)
- Answer c
-
\(-27y^{15}\)
- Answer d
-
\(\dfrac{1}{a^{18}b^{21}}\)
- Answer e
-
\(\dfrac{r^{12}}{s^8}\)
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.
\[(e^{−2}f^2)^7=\dfrac{f^{14}}{e^{14}}\]
Let’s rewrite the original problem differently and look at the result.
\[\begin{align*} (e^{-2}f^2)^7 &= \left(\dfrac{f^2}{e^2}\right)^7\\ &= \dfrac{f^{14}}{e^{14}} \end{align*}\]
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
\[\begin{align*} (e^{-2}f^2)^7 &= \left(\dfrac{f^2}{e^2}\right)^7\\ &= \dfrac{(f^2)^7}{(e^2)^7}\\ &= \dfrac{f^{2\times7}}{e^{2\times7}}\\ &= \dfrac{f^{14}}{e^{14}} \end{align*}\]
For any real numbers a and b and any integer n, the power of a quotient rule of exponents states that
\[\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\]
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. \((m^{-2}n^{-2})^3=\left(\dfrac{1}{m^2n^2}\right)^3=\dfrac{(1)^3}{(m^2n^2)^3}=\dfrac{1}{(m^2)^3(n^2)^3}=\dfrac{1}{m^{2\times3}n^{2\times3}}=\dfrac{1}{m^6n^6}\)
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
- \(\left(\dfrac{b^5}{c}\right)^3\)
- \(\left(\dfrac{5}{u^8}\right)^4\)
- \(\left(\dfrac{-1}{w^3}\right)^{35}\)
- \((p^{-4}q^3)^8\)
- \((c^{-5}d^{-3})^4\)
- Answer a
-
\(\dfrac{b^{15}}{c^3}\)
- Answer b
-
\(\dfrac{625}{u^{32}}\)
- Answer c
-
\(\dfrac{-1}{w^{105}}\)
- Answer d
-
\(\dfrac{q^{24}}{p^{32}}\)
- Answer 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.
Simplify each expression and write the answer with positive exponents only.
- \((6m^2n^{-1})^3\)
- \(17^5\times17^{-4}\times17^{-3}\)
- \(\left(\dfrac{u^{-1}v}{v^{-1}}\right)^2\)
- \((-2a^3b^{-1})(5a^{-2}b^2)\)
- \((x^2\sqrt{2})^4(x^2\sqrt{2})^{-4}\)
- \(\dfrac{(3w^2)^5}{(6w^{-2})^2}\)
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. \[\begin{align*} 17^5\times17^{-4}\times17^{-3} &= 17^{5-4-3} && \text{ The product rule}\\ &= 17^{-2} && \text{ Simplify}\\ &= \dfrac{1}{17^2} \text{ or } \dfrac{1}{289} && \text{ The negative exponent rule} \end{align*}\]
c. \[\begin{align*} \left ( \dfrac{u^{-1}v}{v^{-1}} \right )^2 &= \dfrac{(u^{-1}v)^2}{(v^{-1})^2} && \text{ The power of a quotient rule}\\ &= \dfrac{u^{-2}v^2}{v^{-2}} && \text{ The power of a product rule}\\ &= u^{-2}v^{2-(-2)} && \text{ The quotient rule}\\ &= u^{-2}v^4 && \text{ Simplify}\\ &= \dfrac{v^4}{u^2} && \text{ The negative exponent rule} \end{align*}\]
d. \[\begin{align*} \left (-2a^3b^{-1} \right ) \left(5a^{-2}b^2 \right ) &= \left (x^2\sqrt{2} \right )^{4-4} && \text{ Commutative and associative laws of multiplication}\\ &= -10\times a^{3-2}\times b^{-1+2} && \text{ The product rule}\\ &= -10ab && \text{ Simplify} \end{align*}\]
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. \[\begin{align*} \dfrac{(3w^2)^5}{(6w^{-2})^2} &= \dfrac{(3)^5\times(w^2)^5}{(6)^2\times(w^{-2})^2} && \text{ The power of a product rule}\\ &= \dfrac{3^5w^{2\times5}}{6^2w^{-2\times2}} && \text{ The power rule}\\ &= \dfrac{243w^{10}}{36w^{-4}} && \text{ Simplify}\\ &= \dfrac{27w^{10-(-4)}}{4} && \text{ The quotient rule and reduce fraction}\\ &= \dfrac{27w^{14}}{4} && \text{ Simplify} \end{align*}\]
Access these online resources for additional instruction and practice with exponents and scientific notation.
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. See Example .
- Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See Example .
- Powers of exponential expressions with the same base can be simplified by multiplying exponents. See Example .
- An expression with exponent zero is defined as 1. See Example .
- An expression with a negative exponent is defined as a reciprocal. See Example and Example .
- The power of a product of factors is the same as the product of the powers of the same factors. See Example .
- The power of a quotient of factors is the same as the quotient of the powers of the same factors. See Example .
- The rules for exponential expressions can be combined to simplify more complicated expressions. See Example .