Math 8 · 8.EE.A.1

Powers of powers, products and quotients

A power is itself raised to a power, or a whole product or fraction is. How many factors does that make, and of what?

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Multiplying and dividing powers with the same baseMath 8

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Rational exponentsAlgebra 2

Where you are 3 in Laws of exponents

Number9

Exponents and roots5

Laws of exponents3

Multiplying and dividing powers with the same base

Powers of powers, products and quotients

Zero and negative exponents

Why this exists

Multiplying powers with the same base answers one question: a^m \cdot a^n is a^{m+n}, because two rows of factors line up into one longer row. Powers turn up in other shapes too. A square field whose side is 10^3 meters has area (10^3)^2 square meters: a power, squared. A cube whose edge is 5x has volume (5x)^3: a product, cubed. And the fraction \displaystyle \frac{2}{3} used as a factor four times is \displaystyle \left(\frac{2}{3}\right)^4: a quotient raised to a power.

Multiplying everything out is slow with numbers and impossible with letters, since x has no value. What is wanted is the answer written straight down as a single power, or a product of single powers, and it comes from the same place as the product of powers: an exponent counts factors, so count them. The count gives three rules. The exponents multiply. An exponent outside a product goes to every factor of it. An exponent outside a fraction goes to the top and to the bottom. The same count shows where the rules stop: at stacked exponents, and at an exponent outside a sum.

Definition

For a number a and a positive whole number n, the power a^n is the product of n factors, each equal to a (exponent notation):

a^n = \underbrace{a \cdot a \cdots a}_{n \text{ factors}}

Here a is the base and n is the exponent. A number written with no exponent has exponent 1: a = a^1.

An exponent applies to the one thing written directly before it. When that thing is a bracket, the exponent applies to the whole of what is inside:

2x^3 means 2 \cdot x \cdot x \cdot x. Only x is cubed.
(2x)^3 means (2x)(2x)(2x). The whole product is cubed: a power of a product.
(x^2)^3 means x^2 \cdot x^2 \cdot x^2. A power is used three times as a factor: a power of a power.
\displaystyle \left(\frac{x}{2}\right)^3 means \displaystyle \frac{x}{2} \cdot \frac{x}{2} \cdot \frac{x}{2}: a power of a quotient.

When exponents are stacked with no bracket, as in a^{m^n}, the top exponent belongs to the one just under it. So a^{m^n} means a^{(m^n)}, worked from the top down: 2^{3^2} is 2^9, not (2^3)^2.

Intuition

Count the factors. (a^2)^3 is three copies of a^2, and each copy is two factors of a. Three groups of two make six factors, so (a^2)^3 = a^6.

Figure 1(a^2)^3 is three copies of a^2. Three groups of two factors make 3 \cdot 2 = 6 factors of a.

Compare this with the product of powers. In a^2 \cdot a^3 two different groups are lined up once, so their sizes add: 2 + 3. In (a^2)^3 one group is repeated, three times over, so its size is multiplied: 2 \cdot 3. Adding is for lining groups up; multiplying is for repeating one.

A product inside the bracket works by sorting. (ab)^4 is four copies of ab, which written out is a row of eight factors with a and b taking turns: a \cdot b \cdot a \cdot b \cdot a \cdot b \cdot a \cdot b. The order of factors in a product does not change it, so the a’s can be gathered at one end and the b’s at the other. Four of each: a^4 b^4.

Figure 2(ab)^4 is eight factors, a and b taking turns. Sorted, they are four a’s and four b’s.

A fraction inside the bracket works the same way, because multiplying fractions multiplies the tops together and the bottoms together. Three copies of \displaystyle \frac{a}{b} give three a’s on top and three b’s underneath.

Area gives a picture of the first rule. A square whose side is 3^2 = 9 units long has area (3^2)^2. Ruled into unit squares it has 9 rows of 9, which is 81, and 81 = 3^4. The thick lines count the same squares a second way: the square splits into 3 \cdot 3 = 3^2 blocks, and each block holds 3 \cdot 3 = 3^2 unit squares, so the area is 3^2 \cdot 3^2 = 3^4.

Figure 3A square with side 3^2. Its area is (3^2)^2 = 81 = 3^4, counted as 9 rows of 9 or as 3^2 blocks of 3^2.

Results

Power of a power. For any number a and positive whole numbers m and n,

(a^m)^n = a^{mn}

Power of a product. For any numbers a and b and any positive whole number n,

(ab)^n = a^n b^n

The same holds for three or more factors: (abc)^n = a^n b^n c^n.

Power of a quotient. For any numbers a and b with b \neq 0, and any positive whole number n,

\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

Read from right to left, the power of a product combines powers of different bases that share an exponent: 4^3 \cdot 25^3 = (4 \cdot 25)^3 = 100^3. The exponents must match: 4^3 \cdot 25^2 is 40{,}000, which is not a power of 100 at all.

The rules cover a power and a product of factors. Two other forms give different numbers in general:

Stacked exponents. a^{m^n} has exponent m^n, not mn. For example (3^2)^3 = 3^6 = 729, while 3^{2^3} = 3^8 = 6561.
A power of a sum. (a + b)^n is in general not a^n + b^n. For example (2 + 5)^2 = 7^2 = 49, while 2^2 + 5^2 = 4 + 25 = 29. The square of a sum carries an extra term 2ab, here 2 \cdot 2 \cdot 5 = 20, and indeed 49 = 29 + 20.

Why it is true

Power of a power. By the meaning of the outer exponent, (a^m)^n is n factors, each equal to a^m. The product of powers adds exponents when powers of one base are multiplied, so

\begin{aligned} (a^m)^n &= \underbrace{a^m \cdot a^m \cdots a^m}_{n \text{ factors}} \\ &= a^{\,m + m + \cdots + m} \\ &= a^{mn} \end{aligned}

The exponent on the middle line is n copies of m added together, and that is mn. Since mn = nm, also (a^m)^n = (a^n)^m.

Power of a product. (ab)^n is n factors, each equal to ab, so it is n copies of a and n copies of b in alternating order. The commutative property lets the factors be put in any order, and the associative property lets them be grouped in any way, so the a’s are collected into one group and the b’s into another:

\begin{aligned} (ab)^n &= \underbrace{(ab)(ab)\cdots(ab)}_{n \text{ factors}} \\ &= \underbrace{a \cdots a}_{n} \cdot \underbrace{b \cdots b}_{n} \\ &= a^n b^n \end{aligned}

With a third factor the argument is the same: three letters, sorted into three groups of n.

Power of a quotient. The fraction \displaystyle \frac{a}{b} is a number only when b \neq 0. Taking it as a factor n times and multiplying the fractions, tops with tops and bottoms with bottoms, gives

\begin{aligned} \left(\frac{a}{b}\right)^n &= \underbrace{\frac{a}{b} \cdot \frac{a}{b} \cdots \frac{a}{b}}_{n \text{ factors}} \\ &= \frac{a \cdots a}{b \cdots b} = \frac{a^n}{b^n} \end{aligned}

with n factors on top and n underneath. Since b \neq 0, also b^n \neq 0.

Why stacked exponents differ. The exponent of (a^m)^n is m added n times; the exponent of a^{m^n} is m multiplied n times. For whole numbers m, n \geq 2 these agree only when m = n = 2 (exercise 23), so apart from a few bases such as 0 and 1 the two are different numbers.

Why a sum is different. The argument for (ab)^n sorted the factors of a long product. In (a + b)^2 the bracket holds a sum, and there is nothing to sort. The square means (a + b)(a + b), and the distributive property multiplies it out:

\begin{aligned} (a + b)(a + b) &= a(a + b) + b(a + b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + 2ab + b^2 \end{aligned}

That is a^2 + b^2 plus 2ab, and 2ab is 0 only when a = 0 or b = 0. Higher powers of a sum have extra terms of the same kind, worked out in special products.

Worked examples

Example 1

A power raised to a power

Simplify (a) (x^4)^5 and (b) (6^2)^3, and give (b) as a whole number.

Solution

(a) (x^4)^5 is five copies of x^4, and each copy is four factors of x, so there are 4 \cdot 5 = 20 factors in all:

(x^4)^5 = x^{4 \cdot 5} = x^{20}

(b) The same rule gives (6^2)^3 = 6^{2 \cdot 3} = 6^6 = 46{,}656. Working from the inside out agrees: 6^2 = 36, and 36^3 = 36 \cdot 1296 = 46{,}656.

Example 2

A product raised to a power

Simplify (a) (-2p)^5 and (b) (3m^2n^4)^3.

Solution

(a) The bracket holds the product of -2 and p, so the fifth power goes to each factor:

(-2p)^5 = (-2)^5 p^5 = -32p^5

The coefficient is negative because a product of five negative factors is negative: they pair off into two positive products with one negative factor left over.

(b) The bracket holds three factors, 3, m^2 and n^4. Each is cubed, and the last two are powers of powers, so their exponents multiply:

\begin{aligned} (3m^2n^4)^3 &= 3^3 \, (m^2)^3 \, (n^4)^3 \\ &= 27m^6n^{12} \end{aligned}

A check at m = 2 and n = 1: the left side is (3 \cdot 4 \cdot 1)^3 = 12^3 = 1728, and the right side is 27 \cdot 64 \cdot 1 = 1728. Multiplying whole monomials, coefficients and all, is the subject of multiplying and dividing monomials.

Example 3

A quotient raised to a power

Simplify (a) \displaystyle \left(\frac{2}{7}\right)^3 and (b) \displaystyle \left(\frac{r^2}{s^5}\right)^4, where s \neq 0.

Solution

(a) The cube goes to the top and to the bottom:

\left(\frac{2}{7}\right)^3 = \frac{2^3}{7^3} = \frac{8}{343}

(b) The fourth power goes to the top and to the bottom, and each of those is a power of a power:

\left(\frac{r^2}{s^5}\right)^4 = \frac{(r^2)^4}{(s^5)^4} = \frac{r^8}{s^{20}}

Example 4

Several rules in one expression

Simplify (a) (k^5)^3 \cdot (k^2)^4 and (b) \dfrac{(2c^3)^4}{(4c^5)^2}, where c \neq 0.

Solution

(a) Each bracket is a power of a power, and the exponent outside a bracket is applied before the multiplication between the brackets. So the first factor is k^{15} and the second is k^8. These are two different groups of factors of k, lined up, so multiplying powers of one base adds their exponents:

(k^5)^3 \cdot (k^2)^4 = k^{15} \cdot k^8 = k^{23}

(b) Top and bottom are each a power of a product:

\begin{aligned} (2c^3)^4 &= 2^4 \, (c^3)^4 = 16c^{12} \\ (4c^5)^2 &= 4^2 \, (c^5)^2 = 16c^{10} \end{aligned}

The coefficients cancel, \displaystyle \frac{16}{16} = 1, and dividing powers of one base subtracts the exponents:

\frac{16c^{12}}{16c^{10}} = c^{12 - 10} = c^2

A check at c = 2: the top is (2 \cdot 8)^4 = 16^4 = 65{,}536 and the bottom is (4 \cdot 32)^2 = 128^2 = 16{,}384. Their quotient is 4, which is 2^2.

Example 5

The same number in disguise

Which of these equal 2^6?

(2^3)^2, \quad 2^{3^2}, \quad (2^2)^3, \quad 2^3 \cdot 2^2, \quad 4^3, \quad 8^2

Solution

Write each one as a single power of 2.

(2^3)^2 = 2^{3 \cdot 2} = 2^6. Equal.
2^{3^2} is worked from the top down: 3^2 = 9, so it is 2^9 = 512. Not equal.
(2^2)^3 = 2^{2 \cdot 3} = 2^6. Equal.
2^3 \cdot 2^2 lines up two groups, so the exponents add: 2^{3 + 2} = 2^5 = 32. Not equal.
4 = 2^2, so 4^3 = (2^2)^3 = 2^6. Equal.
8 = 2^3, so 8^2 = (2^3)^2 = 2^6. Equal.

Four of the six equal 2^6 = 64. The last two show a new use: a base that is itself a power can be rewritten in a smaller base, and then powers that looked unrelated can be compared.

Example 6

A square plot measured in meters

A square plot of land is 10^3 meters on a side, which is one kilometer. (a) Find its area in square meters. (b) A larger square plot is 6 \times 10^3 meters on a side. Find its area in square meters, in scientific notation.

Solution

(a) The area of a square is its side squared, so it is a power of a power:

(10^3)^2 = 10^{3 \cdot 2} = 10^6 \text{ m}^2

That is 1{,}000{,}000 square meters: two groups of three zeros make six.

(b) The side is a product, so the square goes to each factor:

\begin{aligned} (6 \times 10^3)^2 &= 6^2 \times (10^3)^2 \\ &= 36 \times 10^6 \text{ m}^2 \end{aligned}

In scientific notation the first factor must be at least 1 and less than 10, so write 36 = 3.6 \times 10 and get 3.6 \times 10^7 \text{ m}^2. A check: the side is 6000 meters, and 6000^2 = 36{,}000{,}000.

Boundaries

The exponents here are positive whole numbers. Every argument above counts factors, and a count needs a positive whole number. For a nonzero base the rules stay true for an exponent of zero and for negative exponents, once those are given their meaning in zero and negative exponents; that meaning is chosen so that the rules keep working.
Fractional exponents need a positive base. The power of a power is what fixes the meaning of fractional exponents, but there it is guaranteed only for a > 0. With a negative base it can fail: \left((-3)^2\right)^{1/2} = 9^{1/2} = 3, while \displaystyle (-3)^{2 \cdot \frac{1}{2}} = (-3)^1 = -3.
A quotient needs a nonzero bottom. \displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} requires b \neq 0, since otherwise neither side is a number.
The exponent goes to factors, never to terms. The power of a product and the power of a quotient are about multiplication and division inside the bracket. With a sum or a difference inside, (a + b)^n and (a - b)^n are in general not a^n + b^n and a^n - b^n. For the square of a sum they agree only when a = 0 or b = 0.
Stacked exponents are not a power of a power. a^{m^n} has exponent m^n, and (a^m)^n has exponent mn.
Different bases combine only with equal exponents. a^3 b^2 is not a single power of ab.

Where this goes next

The three rules here, with the product and quotient of powers, make up the laws of exponents. Exponents of 0 and -1 have no count of factors behind them, so their meaning is chosen to keep these laws true, in zero and negative exponents. The power of a power then decides what a fractional exponent must be: if \left(a^{1/2}\right)^2 = a^1 = a, then a^{1/2} is the square root of a, the start of fractional exponents. The power of a product is what lets scientific notation handle powers of large numbers, and repeated growth in exponential functions is powers of powers again: growth by a^m in one period, repeated for n periods, is growth by a^{mn}.

Exercises

Powers of powers

Simplify (y^6)^3.
Simplify (5^2)^4 and give the answer as a whole number.
Simplify (t^9)^7.
Simplify \left((q^2)^3\right)^4.

Products raised to a power

Simplify (4z)^3.
Simplify (-3z)^4.
Simplify (-5g^3)^3.
Simplify (2a^5b)^6.
Simplify (10c^2d^3)^4.

Quotients raised to a power

Simplify \displaystyle \left(\frac{3}{4}\right)^4.
Simplify \displaystyle \left(\frac{n}{6}\right)^2.
Simplify \displaystyle \left(\frac{2v^3}{w}\right)^5, where w \neq 0.
Simplify \displaystyle \left(\frac{-x^4}{5y^2}\right)^3, where y \neq 0.

Several rules together

Simplify (h^5)^2 \cdot (h^3)^3.
Simplify (3p^2)^3 \cdot (p^4)^2.
Simplify \dfrac{(10m^5)^2}{(5m^3)^2}, where m \neq 0, in two ways: once by squaring top and bottom first, and once by simplifying inside a single bracket first.
Write 9^5 as a power of 3, and 8^4 as a power of 2.
Which is larger, 2^{30} or 3^{20}? Decide without working out either number in full.

In context

Joaquina has a square tile whose side is 2^5 millimeters long. Find its area in square millimeters, as a power of 2 and as a whole number.
Hafsa enlarges a photo so that every length in it is 3 times as long, then enlarges the enlarged photo in the same way. By what number has the area of the photo been multiplied? Get the answer in two ways, once with the power of a power and once with the product of powers.
A cube has edges 3 \times 10^2 millimeters long. Find its volume in cubic millimeters, as a whole number and in scientific notation.

Reasoning

Is (a^m)^n always equal to (a^n)^m? Explain.
Evaluate (2^2)^3 and 2^{2^3}. Then show that (a^2)^2 = a^{2^2} for every a, and explain why m = n = 2 is the only pair of whole numbers, both 2 or more, with mn = m^n.
Evaluate (1 + 2)^3 and 1^3 + 2^3. Then show that (a + b)^2 = a^2 + b^2 holds exactly when a = 0 or b = 0.
Prove that (abc)^n = a^n b^n c^n for every positive whole number n, using the rule for two factors.

Answers

y^{18}. Three copies of six factors: 6 \cdot 3 = 18.
5^8 = 390{,}625. The exponents multiply, 2 \cdot 4 = 8; and 5^4 = 625, so 5^8 = 625^2 = 390{,}625.
t^{63}, since 9 \cdot 7 = 63.
q^{24}. The inner bracket is q^6, and (q^6)^4 = q^{24}.
64z^3. Each factor is cubed: 4^3 z^3.
81z^4. (-3)^4 = 81, positive because four negative factors pair off.
-125g^9. (-5)^3 = -125 and (g^3)^3 = g^9.
64a^{30}b^6. 2^6 = 64, (a^5)^6 = a^{30} and b^6.
10{,}000c^8d^{12}. 10^4 = 10{,}000, (c^2)^4 = c^8 and (d^3)^4 = d^{12}.
\displaystyle \frac{81}{256}. Top and bottom to the fourth: \displaystyle \frac{3^4}{4^4}.
\displaystyle \frac{n^2}{36}. Top and bottom squared.
\displaystyle \frac{32v^{15}}{w^5}. The top is 2^5 (v^3)^5 = 32v^{15} and the bottom is w^5.
\displaystyle -\frac{x^{12}}{125y^6}. The top is (-1)^3 (x^4)^3 = -x^{12} and the bottom is 5^3 (y^2)^3 = 125y^6.
h^{19}. The brackets give h^{10} and h^9, and lining them up adds the exponents.
27p^{14}. (3p^2)^3 = 27p^6 and (p^4)^2 = p^8; then p^6 \cdot p^8 = p^{14}.
4m^4 both ways. Squaring first: \displaystyle \frac{100m^{10}}{25m^6} = 4m^4. Simplifying inside first: \displaystyle \frac{10m^5}{5m^3} = 2m^2, so the expression is \displaystyle \left(\frac{10m^5}{5m^3}\right)^2 = (2m^2)^2 = 4m^4, using the power of a quotient read from right to left.
9^5 = 3^{10} and 8^4 = 2^{12}. Since 9 = 3^2, 9^5 = (3^2)^5 = 3^{10}; since 8 = 2^3, 8^4 = (2^3)^4 = 2^{12}.
3^{20} is larger. 2^{30} = (2^3)^{10} = 8^{10} and 3^{20} = (3^2)^{10} = 9^{10}, and ten factors of 9 make more than ten factors of 8.
2^{10} = 1024 square millimeters. The area is (2^5)^2 = 2^{10}.
By 81. After both enlargements every length is 3 \cdot 3 = 3^2 times as long, so the area is multiplied by (3^2)^2 = 3^4 = 81. Or: each enlargement multiplies the area by 3^2, and two of them multiply it by 3^2 \cdot 3^2 = 3^4 = 81.
27{,}000{,}000 cubic millimeters, which is 2.7 \times 10^7. The volume is (3 \times 10^2)^3 = 3^3 \times (10^2)^3 = 27 \times 10^6.
Yes. (a^m)^n = a^{mn} and (a^n)^m = a^{nm}, and mn = nm because multiplication of numbers does not depend on order.
(2^2)^3 = 2^6 = 64 but 2^{2^3} = 2^8 = 256. For the pair 2, 2: (a^2)^2 = a^4 and a^{2^2} = a^4, so they agree. For the rest, write m^n = m \cdot m^{n-1}. Since m \geq 2, m^{n-1} \geq 2^{n-1}, and 2^{n-1} \geq n for every n \geq 2 (doubling from 2 keeps pace with counting up by one), so m^n \geq m \cdot n. Equality needs both steps to be equalities: m^{n-1} = 2^{n-1} forces m = 2, and 2^{n-1} = n forces n = 2.
(1 + 2)^3 = 27 and 1^3 + 2^3 = 9. For the square, (a + b)^2 = a^2 + 2ab + b^2 by the distributive property, so (a + b)^2 = a^2 + b^2 exactly when 2ab = 0. A product is 0 exactly when one of its factors is 0, so this happens exactly when a = 0 or b = 0.
The product abc is (ab)c, by the associative property. Applying the rule for two factors to ab and c gives \left((ab)c\right)^n = (ab)^n c^n, and applying it again to (ab)^n gives a^n b^n c^n.