2.6: Rules of Exponents
- Page ID
- 49350
Overview
- The Product Rule for Exponents
- The Quotient Rule for Exponents
- Zero as an Exponent
We will begin our study of the rules of exponents by recalling the definition of exponents.
If \(x\) is any real number and \(n\) is a natural number, then
\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\)
An exponent records the number of identical factors in a multiplication.
In \(x^n\),
\(x\) is the base
\(n\) is the exponent
The number represented by \(x^n\) is called a power
The term \(x^n\) is read as "\(x\) to the \(n\)th."
The Product Rule for Exponents
The first rule we wish to develop is the rule for multiplying two exponential quantities having the same base and natural number exponents. The following examples suggest this rule:
\(
\begin{aligned}
&\begin{array}{c}
x^{2} \cdot x^{4}=\underbrace{x x}_{2} \cdot \underbrace{x x x x}_{4}=\underbrace{x x x x x x}_{6}=x^{6} \\
2+4=6
\end{array}\\
\end{aligned}
\)
\(
a \cdot a^{2}=\underbrace{a}_{1} \cdot \underbrace{aa}_{2}=\underbrace{a a a}_{3}=a^{3} \\
1 + 2 = 3
\)
If \(x\) is a real number and \(n\) and \(m\) are natural numbers,
\(x^nx^m = x^{n+m}\)
To multiply two exponential quantities having the same base, add the exponents. Keep in mind that the exponential quantities being multiplied must have the same base for this rule to apply.
Sample Set A
Find the following products. All exponents are natural numbers.
\(x^3 \cdot x^5 = x^{3+5} = x^8\)
\(a^6 \cdot a^14 = a^{6+14} = a^20\)
\(y^5 \cdot y = y^5 \cdot y^1 = y^{5+1} = y^6\)
\((x-2y)^8(x-2y)^5 = (x-2y)^{8+5} = (x-2y)^{13}\)
\(x^3y^4 \not = (xy)^{3+4}\)
Since the bases are not the same, the product rule does not apply.
Practice Set A
Find each product.
\(x^2 \cdot x^5\)
- Answer
-
\(x^{2+5} = x^7\)
\(x^9 \cdot x^4\)
- Answer
-
\(x^{9+4} = x^{13}\)
\(y^6 \cdot y^4\)
- Answer
-
\(y^{6+4} = y^{10}\)
\(c^12 \cdot c^8\)
- Answer
-
\(c^{12+8} = c^{20}\)
\((x+2)^3 \cdot (x+2)^5\)
- Answer
-
\((x+2)^{3+5} = (x+2)^8\)
Sample Set B
We can use the first rule of exponents (and the others that we will develop) along with the properties of real numbers.
\(2x^3 \cdot 7x^5 = 2 \cdot 7 \cdot x^{3+5} = 14x^8\)
We used the commutative and associative properties of mulitplication. In practice, we use these properties "mentally" (as signiffied by the drawing of the box). We don't actually write the second step.
\(4y^3 \cdot 6y^2 = 4 \cdot 6 \cdot y^{3+2} = 24y^5\)
\(9a^2b^6(8ab^42b^3) = 9 \cdot 8 \cdot 2a^{2+1}b^{6+4+3} = 144a^3b^{13}\)
\(5(a+6)^2 \cdot 3(a+6)^8 = 5 \cdot 3(a+6)^{2+8} = 15(a+6)^{10}\)
\(4x^3 \cdot 12 \cdot y^2 = 48x^3y^2\)
Practice Set B
Perform each multiplication in one step.
\(3x^5 \cdot 2x^2\)
- Answer
-
\(6x^7\)
\(6y^3 \cdot 3y^4\)
- Answer
-
\(18y^7\)
\(4a^3b^2 \cdot 9a^2b\)
- Answer
-
\(36a^5b^3\)
\(x^4 \cdot 4y^2 \cdot 2x^2 \cdot 7y^6\)
- Answer
-
\(56x^6y^8\)
\((x-y)^3 \cdot 4(x-y)^2\)
- Answer
-
\(4(x-y)^5\)
\(8x^4y^2xx^3y^5\)
- Answer
-
\(8x^8y^7\)
\(2aaa^3(ab^2a^3)b6ab^2\)
- Answer
-
\(12a^{10}b^5\)
\(a^n \cdot a^m \cdot a^r\)
- Answer
-
\(a^{n+m+r}\)
The Quotient Rule for Exponents
The second rule we wish to develop is the rule for dividing two exponential quantities having the same base and natural number exponents.
The following examples suggest a rule for dividing two exponential quantities having the same base and natural number exponents.
\(
\dfrac{x^{5}}{x^{2}}=\dfrac{x x x x x}{x x}=\dfrac{\cancel{(x x)} x x x}{\cancel{(x x)}}=x x x=x^{3} . \text { Notice that } 5-2=3
\)
\(
\dfrac{a^8}{a^3} = \dfrac{aaaaaaaa}{aaa} = \dfrac{(\cancel{aaa})aaaaa}{(\cancel{aaa})} = aaaaa = a^5. \text { Notice that } 8-3=5
\)
If \(x\) is a real number and \(n\) and \(m\) are natural numbers,
\(\dfrac{x^n}{x^m} = x^{n-m}, x \not = 0.\)
Sample Set C
Find the following quotients. All exponents are natural numbers.
\(\dfrac{x^5}{x^2} = x^{5-2} = x^3\)
\(\dfrac{27a^3b^6c^2}{3a^2bc} = 9a^{3-2}b^{6-1}c^{2-1} = 9ab^5c\)
\(\dfrac{15x^{□}}{3x^{△}} = 5x^{□-△}\)
The bases are the same, so we subtract the exponents. Although we don’t know exactly what □−△ is, the notation □−△ indicates the subtraction.
Practice Set C
Find each quotient
\(\dfrac{y^9}{y^5}\)
- Answer
-
\(y^4\)
\(\dfrac{a^7}{a}\)
- Answer
-
\(a^6\)
\(\dfrac{(x+6)^5}{(x+6)^3}\)
- Answer
-
\((x+6)^2\)
\(\dfrac{26x^4y^6z^2}{13x^2y^2z}\)
- Answer
-
\(2x^2y^4z\)
When we make the subtraction, \(n-m\), in the division, \(\dfrac{x^n}{x^m}\), there are three possibilities for the values of the exponents:
The exponent of the numerator is greater than the exponent of the denominator, that is, \(n > m\). Thus, the exponent, \(n-m\), is a natural number.
The exponents are the same, that is, \(n = m\). Thus, the exponent, \(n - m\), is zero, a whole number.
The exponent of the denominator is greater than the exponent of the numerator, that is, \(n < m\). Thus, the exponent, \(n - m\), is an integer.
Zero as an Exponent
In Sample Set C, the exponents of the numerators were greater than the exponents of the denominators. Let’s study the case when the exponents are the same.
When the exponents are the same, say \(n\), the subtraction \(n-n\) produces \(0\).
Thus, by the second rule of exponents, \(\dfrac{x^n}{x^n} = x^{n-n} = x^0\)
But what real number, if any, does \(x^0\) represent? Let's think for a moment about our experience with division in arithmetic. We know that any nonzero number divided by itself is one.
\(\dfrac{8}{8} = 1\), \(\dfrac{43}{43} = 1\), \(\dfrac{258}{258} = 1\)
Since the letter \(x\) represents some nonzero real number divided by itself. Then \(\dfrac{x^n}{x^n} = 1\).
But we have also established that if \(x \not = 0\), \(\dfrac{x^n}{x^n} = x^0\). We now have that \(\dfrac{x^n}{x^n} = x^0\) and \(\dfrac{x^n}{x^n} = 1\). This implies that \(x^0 = 1\), \(x \not = 0\).
Exponents can now be natural numbers and zero. We have enlarged our collection of numbers that can be used as exponents from the collection of natural numbers to the collection of whole numbers.
If \(x \not = 0\), \(x^0 = 1\)
Any number, other than \(0\), raised to the power of \(0\), is \(1\). \(0^0\) has no meaning (it does not represent a number).
Sample Set D
Find each value. Assume the base is not zero.
\(6^0 = 1\)
\(246^0 = 1\)
\((2a+5)^0 = 1\)
\(4y^0 = 4 \cdot 1 = 4\)
\(\dfrac{y^6}{y^6} = y^0 = 1\)
\(\dfrac{2x^2}{x^2} = 2x^0 = 2 \cdot 1 = 2\)
\(
\begin{aligned}
\dfrac{5(x+4)^{8}(x-1)^{5}}{5(x+4)^{3}(x-1)^{5}} &=(x+4)^{8-3}(x-1)^{5-5} \\
&=(x+4)^{5}(x-1)^{0} \\
&=(x+4)^{5}
\end{aligned}
\)
Practice Set D
Find each value. Assume the base is not zero.
\(\dfrac{y^7}{y^3}\)
- Answer
-
\(y^{7-3} = y^4\)
\(\dfrac{6x^4}{2x^3}\)
- Answer
-
\(3x^{4-3} = 3x\)
\(\dfrac{14a^7}{7a^2}\)
- Answer
-
\(2a^{7-2} = 2a^5\)
\(\dfrac{26x^2y^5}{4xy^2}\)
- Answer
-
\(\dfrac{13}{2}xy^3\)
\(\dfrac{36a^4b^3c^8}{8ab^3c^6}\)
- Answer
-
\(\dfrac{9}{2}a^3c^2\)
\(\dfrac{51(a-4)^3}{17(a-4)}\)
- Answer
-
\(3(a-4)^2\)
\(\dfrac{52a^7b^3(a+b)^8}{26a^2b(a+b)^8}\)
- Answer
-
\(2a^5b^2\)
\(\dfrac{a^n}{a^3}\)
- Answer
-
\(a^{n-3}\)
\(\dfrac{14x^ry^pz^q}{2x^ry^hz^5}\)
- Answer
-
\(7y^{p-h}z^{q-5}\)
Exercises
Use the product rule and quotient rule of exponents to simplify the following problems. Assume that all bases are nonzero and that all exponents are whole numbers.
\(3^2 \cdot 3^3\)
- Answer
-
\(3^5 = 243\)
\(5^2 \cdot 5^4\)
\(9^0 \cdot 9^2\)
- Answer
-
\(9^2 = 81\)
\(7^3 \cdot 7^0\)
\(2^4 \cdot 2^5\)
- Answer
-
\(2^9 = 512\)
\(x^5x^4\)
\(x^2x^3\)
- Answer
-
\(x^5\)
\(a^9a^7\)
\(y^5y^7\)
- Answer
-
\(y^12\)
\(m^{10}m^2\)
\(k^8k^3\)
- Answer
-
\(k^{11}\)
\(y^3y^4y^6\)
\(3x^2 \cdot 2x^5\)
- Answer
-
\(6x^7\)
\(a^2a^3a^8\)
\(4y^4 \cdot 5y^6\)
- Answer
-
\(20y^{10}\)
\(2a^3b^2 \cdot 3ab\)
\(12xy^3z^2 \cdot 4x^2y^2z \cdot 3x\)
- Answer
-
\(144x^4y^5z^3\)
\((3ab)(2a^2b)\)
\((4x^2)(8xy^3)\)
- Answer
-
\(32x^3y^3\)
\((2xy)(3y)(4x^2y^5)\)
\((\dfrac{1}{4}a^2b^4)(\dfrac{1}{2}b^4)\)
- Answer
-
\(\dfrac{1}{8}a^2b^8\)
\((\dfrac{3}{8})(\dfrac{16}{21}x^2y^3)(x^3y^2)\)
\(\dfrac{8^5}{8^3}\)
- Answer
-
\(8^2 = 64\)
\(\dfrac{6^4}{6^3}\)
\(\dfrac{2^9}{2^4}\)
- Answer
-
\(2^5 = 32\)
\(\dfrac{4^{16}}{4^{13}}\)
\(\dfrac{x^5}{x^3}\)
- Answer
-
\(x^2\)
\(\dfrac{y^4}{y^3}\)
\(\dfrac{y^9}{y^4}\)
- Answer
-
\(y^5\)
\(\dfrac{k^{16}}{k^{13}}\)
\(\dfrac{x^4}{x^2}\)
- Answer
-
\(x^2\)
\(\dfrac{y^5}{y^2}\)
\(\dfrac{m^{16}}{m^9}\)
- Answer
-
\(m^7\)
\(\dfrac{a^9b^6}{a^5b^2}\)
\(\dfrac{y^3w^{10}}{yw^5}\)
- Answer
-
\(y^2w^5\)
\(\dfrac{m^{17}n^{12}}{m^{16}n^{10}}\)
\(\dfrac{x^5y^7}{x^3y^4}\)
- Answer
-
\(x^2y^3\)
\(\dfrac{15x^{20}y^{24}z^4}{5x^{19}yz}\)
\(\dfrac{e^{11}}{e^{11}}\)
- Answer
-
\(e^0 = 1\)
\(\dfrac{6r^4}{6r^4}\)
\(\dfrac{x^0}{x^0}\)
- Answer
-
\(x^0 = 1\)
\(\dfrac{a^0b^0}{c^0}\)
\(\dfrac{8a^4b^0}{4a^3}\)
- Answer
-
\(2a\)
\(\dfrac{24x^4y^4z^0w^8}{9xyw^7}\)
\(t^2(y^4)\)
- Answer
-
\(t^2y^4\)
\(x^3(\dfrac{x^6}{x^3})\)
\(a^4b^6(\dfrac{a^{10}b^{16}}{a^5b^7})\)
- Answer
-
\(a^9b^{15}\)
\(3a^2b^3(\dfrac{14a^2b^5}{2b})\)
\(\dfrac{(x+3y)^{11}(2x-1)^4}{(x+3y)^3(2x-1)}\)
- Answer
-
\((x+3y)^8(2x-1)^3\)
\(\dfrac{40 x^{5} z^{10}\left(z-x^{4}\right)^{12}(x+z)^{2}}{10 z^{7}\left(z-x^{4}\right)^{5}}\)
\(x^nx^r\)
- Answer
-
\(x^{n+r}\)
\(a^xb^yz^{5z}\)
\(x^n \cdot x^{n+3}\)
- Answer
-
\(x^{2n+3}\)
\(\dfrac{x^{n+3}}{x^n}\)
\(\dfrac{x^{n+2}x^3}{x^4x^n}\)
- Answer
-
\(x\)
Exercises for Review
What natural numbers can replace \(x\) so that the statement \(-5 < x \le 3\) is true?
Use the distributive property to expand \(4x(2a + 3b)\)
- Answer
-
\(8ax + 12bx\)
Express \(xxxyyyy(a+b)(a+b)\) using exponents.
Find the value of \(4^2 + 3^2 \cdot 2^3 - 10 \cdot 8\)
- Answer
-
\(8\)
Find the value of \(\dfrac{4^2+(3+2)^2-1}{2^3 \cdot 5} + \dfrac{2^4(3^2-2^3+}{4^2}\).