Unit 3, lesson 20

Exponent Rules

How the rules for products, quotients, powers, zero and negative exponents follow from counting repeated factors, and which values of the base each rule must exclude.

  • Theory part 39
  • Practice part 40
  • Powers and Expressions

Theory, part 39

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Practice, part 40

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Fold a sheet of paper in half and it is 2 layers thick. Fold again and it is 4, then 8, then 16. After 5 folds it has 2⋅2⋅2⋅2⋅2=322 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32 layers, which is written 252^5. Every exponent rule comes from counting factors like these, so if you forget a rule you can always rebuild it by writing a small power out in full. On the SAT these rules decide which of four expressions is equivalent to a given one.

See also Powers, Roots, and Signs in Foundations for Calculus and Using Exponent and Logarithm Laws in Foundations for Calculus.

Key terms

Power, base and exponent

Definition For a real number aa and a positive integer nn, the power ana^n is the product of nn factors equal to aa,

an=a⋅a⋯a⏟n factors.a^n = \underbrace{a \cdot a \cdots a}_{n \text{ factors}}.

The number aa is the base and nn is the exponent. The exponent applies only to the symbol directly below it, so −32=−(32)=−9-3^2 = -(3^2) = -9, while (−3)2=9(-3)^2 = 9, and 2x32x^3 means 2⋅x32 \cdot x^3, not (2x)3(2x)^3.

In plain words The exponent counts how many times the base is used as a factor. Parentheses decide what the base is.

Example Here 53=5⋅5⋅5=1255^3 = 5 \cdot 5 \cdot 5 = 125 and (−2)4=16(-2)^4 = 16, but −24=−16-2^4 = -16. With a variable, 3x23x^2 at x=4x = 4 is 3⋅16=483 \cdot 16 = 48, while (3x)2(3x)^2 at x=4x = 4 is 122=14412^2 = 144.

Product rule for powers

Definition For a real number aa and positive integers mm and nn,

am⋅an=am+n.a^m \cdot a^n = a^{m + n}.

The rule also holds for all integers mm and nn when a≠0a \ne 0, using the definitions of zero and negative exponents below. It needs the same base on both factors.

In plain words Multiplying powers of the same base just adds up the factors, so you add the exponents.

Example Write it out to see why, x2⋅x3=(x⋅x)(x⋅x⋅x)=x5x^2 \cdot x^3 = (x \cdot x)(x \cdot x \cdot x) = x^5. With coefficients, multiply the numbers and add the exponents of each variable,

(4x3y)(5x2y4)=20x5y5.(4x^3 y)(5x^2 y^4) = 20x^5 y^5.

A product with different bases, such as 23⋅32=8⋅9=722^3 \cdot 3^2 = 8 \cdot 9 = 72, does not combine into one power.

Quotient rule for powers

Definition For a real number a≠0a \ne 0 and integers mm and nn (zero and negative exponents are defined below),

aman=am−n.\frac{a^m}{a^n} = a^{m - n}.

An expression simplified this way keeps the restriction of the original. For example x5x2\frac{x^5}{x^2} is undefined at x=0x = 0, so it equals x3x^3 only for x≠0x \ne 0.

In plain words Matching factors on top and bottom cancel in pairs, and the exponents subtract. The excluded value stays excluded even after it no longer shows.

Example Cancel factor by factor,

x5x2=x⋅x⋅x⋅x⋅xx⋅x=x3for x≠0.\begin{gathered} \frac{x^5}{x^2} = \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x} = x^3 \\[0.4em] \text{for } x \ne 0. \end{gathered}

At x=2x = 2 the left side is 324=8\frac{32}{4} = 8 and the right side is 23=82^3 = 8. At x=0x = 0 the left side is undefined, though x3x^3 alone would give 00.

Power of a power and power of a product

Definition For a real number aa and positive integers mm and nn,

(am)n=amn.(a^m)^n = a^{mn}.

For real numbers aa and bb and a positive integer nn,

(ab)n=anbn,(ab)n=anbn with b≠0.\begin{gathered} (ab)^n = a^n b^n, \\[0.4em] \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \text{ with } b \ne 0. \end{gathered}

These also hold for all integers when the bases involved are nonzero. They apply to products and quotients only. A power of a sum is not the sum of the powers, so in general (a+b)2≠a2+b2(a + b)^2 \ne a^2 + b^2.

In plain words An outside exponent multiplies each exponent inside and reaches every factor of a product. It does not split across a plus sign.

Example Raise each factor to the third power,

(2x3y2)3=23x9y6=8x9y6.(2x^3 y^2)^3 = 2^3 x^9 y^6 = 8x^9 y^6.

For the sum, (3+4)2=49(3 + 4)^2 = 49, but 32+42=253^2 + 4^2 = 25. The square of a sum also contains the cross term, (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.

Zero exponent

Definition For every real number a≠0a \ne 0,

a0=1.a^0 = 1.

This is the only choice that keeps the quotient rule true, since anan=1\frac{a^n}{a^n} = 1 and also anan=an−n=a0\frac{a^n}{a^n} = a^{n - n} = a^0. The expression 000^0 is left undefined here, so a statement like x0=1x^0 = 1 carries the condition x≠0x \ne 0.

In plain words Any nonzero number to the power zero is 11, not 00. The exponent applies only to its own base.

Example Here 70=17^0 = 1 and (−4)0=1(-4)^0 = 1, but −40=−(40)=−1-4^0 = -(4^0) = -1. For x≠0x \ne 0 the expression 3x03x^0 equals 3⋅1=33 \cdot 1 = 3, while (3x)0=1(3x)^0 = 1.

Negative exponent

Definition For a real number a≠0a \ne 0 and a positive integer nn,

a−n=1an.a^{-n} = \frac{1}{a^n}.

A negative exponent never makes a number negative. It stands for a reciprocal. In particular a−1=1aa^{-1} = \frac{1}{a}, and for a≠0a \ne 0 and b≠0b \ne 0, (ab)−n=(ba)n\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n. As with sums of powers, (a+b)−1(a + b)^{-1} is not a−1+b−1a^{-1} + b^{-1}.

In plain words A negative exponent moves the power to the other side of the fraction bar. The base cannot be 00, because 10\frac{1}{0} is undefined.

Example Here 2−3=182^{-3} = \frac{1}{8} and (25)−2=254\left(\frac{2}{5}\right)^{-2} = \frac{25}{4}. The product rule still works, x−2⋅x5=x3x^{-2} \cdot x^5 = x^3 for x≠0x \ne 0. Watch the sum warning with numbers. Since 2−1+2−1=12^{-1} + 2^{-1} = 1 while (2+2)−1=14(2 + 2)^{-1} = \frac{1}{4}, the two expressions are different.

Here is a typical simplification with positive xx and yy,

(2x3y−2)34x5y=8x9y−64x5y=2x4y−7=2x4y7.\begin{aligned} \frac{(2x^3 y^{-2})^3}{4x^5 y} &= \frac{8x^9 y^{-6}}{4x^5 y} \\[0.4em] &= 2x^4 y^{-7} = \frac{2x^4}{y^7}. \end{aligned}

Check yourself

  1. Which expression is equivalent to (3x2)39x4\dfrac{(3x^2)^3}{9x^4} for x≠0x \ne 0? (A) x2x^2 (B) 3x23x^2 (C) 3x3x (D) x23\dfrac{x^2}{3}
Show the answer

The answer is (B). The numerator is (3x2)3=27x6(3x^2)^3 = 27x^6, and dividing by 9x49x^4 gives 3x23x^2. Choice (A) comes from writing 333^3 as 99, which gives 9x69x4\frac{9x^6}{9x^4}. Choice (D) comes from forgetting to cube the 33, which gives 3x69x4\frac{3x^6}{9x^4}. Choice (C) adds the exponents 2+32 + 3 instead of multiplying them, which gives 27x59x4=3x\frac{27x^5}{9x^4} = 3x.

  1. If 32x⋅9=3103^{2x} \cdot 9 = 3^{10}, what is the value of xx?
Show the answer

The answer is 44. Write 99 as 323^2, so the product rule gives 32x+2=3103^{2x + 2} = 3^{10}. Equal powers of the same base 33 have equal exponents, so 2x+2=102x + 2 = 10 and x=4x = 4. As a check, 38⋅9=6561⋅9=59049=3103^8 \cdot 9 = 6561 \cdot 9 = 59049 = 3^{10}.

  1. For x≠0x \ne 0, what is the value of (5x2)0+5x0(5x^2)^0 + 5x^0? (A) 00 (B) 22 (C) 66 (D) 1010
Show the answer

The answer is (C). The whole product 5x25x^2 is the base of the first power, so (5x2)0=1(5x^2)^0 = 1. In 5x05x^0 the exponent applies only to xx, so it equals 5⋅1=55 \cdot 1 = 5, and the sum is 66. Choice (B) treats 5x05x^0 as (5x)0(5x)^0, choice (D) treats (5x2)0(5x^2)^0 as 55, and choice (A) reads a zero exponent as giving 00.

How this lesson connects