Chapter 10 — The Laws of Exponents

Standard: A.EO.3 (a, b)

A.EO.3 — verbatim. The student will derive and apply the laws of exponents. Students will demonstrate the following Knowledge and Skills: a) Derive the laws of exponents through explorations of patterns, to include products, quotients, and powers of bases. b) Simplify multivariable expressions and ratios of monomial expressions in which the exponents are integers, using the laws of exponents.

By the end of this chapter you will be able to:

• Derive the product law $a^m \cdot a^n = a^{m+n}$ by writing out the factors and counting them, rather than by being told (A.EO.3a)
• Derive the quotient law $\dfrac{a^m}{a^n} = a^{m-n}$ by matching factors off in pairs (A.EO.3a)
• Show that $a^0 = 1$ is forced by a descending pattern and by reading one quotient two ways (A.EO.3a)
• Show that $a^{-n} = \dfrac{1}{a^n}$ is forced by continuing that same pattern below zero (A.EO.3a)
• Derive the three laws for powers of bases — $(a^m)^n = a^{mn}$, $(ab)^n = a^n b^n$, and $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$ — by counting groups of factors (A.EO.3a)
• Tell $-a^n$ apart from $(-a)^n$, and say which symbol the exponent actually sits on (A.EO.3a)
• Simplify multivariable expressions with integer exponents, base by base, leaving no negative exponent in a final answer (A.EO.3b)
• Simplify ratios of monomial expressions with integer exponents, dividing coefficients and subtracting exponents (A.EO.3b)
• Use powers of ten to compare orders of magnitude in real quantities, and check a power on a calculator (A.EO.3b)

Lessons: 10.1 Powers and the Product Law · 10.2 The Quotient Law · 10.3 Zero and Negative Exponents · 10.4 Powers of Bases · 10.5 Multivariable Expressions and Ratios of Monomials

Why this chapter matters. An exponent is a counting device: $a^5$ records that five copies of $a$ are being multiplied. Every law in this chapter is therefore a fact about counting, and that is why none of them has to be memorized. If you can write out the factors, you can rebuild the law on the spot. That matters immediately — polynomial multiplication in Chapter 12 is the product law applied term by term, simplest radical form in Chapter 11 leans on powers of a product, and exponential models in Chapter 17 are built on the same $b^x$ whose behavior at $x = 0$ and at negative $x$ you settle here. It also matters outside mathematics: populations, distances, file sizes, and the doubling of a dividing cell are all reported as powers, and comparing two of them is a subtraction of exponents.

Scope note. This chapter derives the laws of exponents and applies them to expressions. It stays inside integer exponents, because A.EO.3b says so: zero and negative exponents are in scope, and rational or fractional exponents are not. Radicals, and rational exponents limited to $\tfrac12$ and $\tfrac13$, are A.EO.4 in Chapter 11. Adding, subtracting, and multiplying polynomials is A.EO.2 a and b in Chapter 12, factoring is A.EO.2c in Chapter 13, and dividing a polynomial by a monomial or a binomial is A.EO.2d in Chapter 14 — this chapter divides one monomial by another and stops there. Exponential functions $y = ab^x$, their graphs, their domains and ranges, and their transformations are A.F.2 e and f in Chapter 17; nothing here graphs a power or asks how it grows. No figure in this chapter is a graph of a function, on purpose.

Conventions this chapter fixes.

• In $a^n$, the number $a$ is the base, the number $n$ is the exponent, and the whole expression is a power. Read $a^n$ as "$a$ to the $n$th power."
• The exponent reaches only the symbol it sits on. In $-3^2$ the exponent sits on the $3$, so the expression is $-(3 \cdot 3) = -9$. In $(-3)^2$ the parentheses make $-3$ the base, so the expression is $(-3)(-3) = 9$. In $3x^2$ the exponent sits on the $x$ only; in $(3x)^2$ it reaches both.
• $a^1 = a$. One copy of $a$ multiplied together is $a$. An exponent of $1$ is almost never written, but it is always there, and reading $b$ as $b^1$ is what makes the product law work on it.
• Every quotient law carries a restriction. $\dfrac{a^m}{a^n}$ requires $a \neq 0$, because the denominator would otherwise be zero. So do $a^0 = 1$ and $a^{-n} = \dfrac{1}{a^n}$, both of which are derived from a quotient. This chapter states the restriction rather than hiding it, and $0^0$ is left undefined.
• A negative exponent is not a negative number. $2^{-3} = \tfrac18$, a positive number less than one. The minus sign in the exponent says reciprocal, not opposite.
• Final answers use positive exponents, unless an item asks for something else. So $\dfrac{x^3}{x^7}$ is reported as $\dfrac{1}{x^4}$, not as $x^{-4}$, even though the two are equal at every allowed value of $x$.
• A monomial is a number, a variable, or a product of numbers and variables. A ratio of monomials is one monomial divided by another, and simplifying one is the whole content of Lesson 10.5.
• Item numbering runs straight through the chapter, from 1 in Lesson 10.1 to 126 at the end of the review. It does not restart at each lesson.

Calculator note. Algebra 1 has no no-calculator standards, and the Desmos Virginia calculator is available for the entire End-of-Course test. Use it here the way this volume uses it everywhere: to confirm a result you already produced. Entering $2^{-3}$ and seeing $0.125$ is a good habit, because it is one keystroke and it catches the student who thought the answer would be $-8$ or $-6$. What a calculator cannot do is simplify $\dfrac{15x^5y^3}{5x^2y^7}$, since that has no numeric value until $x$ and $y$ are chosen — which is exactly why substituting a couple of values into both the original and your answer is the algebraic version of the same check. The powers worth knowing on sight are the powers of two up to $2^{10} = 1024$ and the powers of ten in both directions.

Lesson 10.1 — Powers and the Product Law

What an exponent counts

A power is shorthand for repeated multiplication:

$$a^5 = a \cdot a \cdot a \cdot a \cdot a$$

The base $a$ is the thing being multiplied, and the exponent $5$ counts how many copies appear. That is the whole definition, and every law in this chapter is a consequence of it.

Two small readings follow immediately.

• $a^1 = a$. One copy of $a$ is $a$. The exponent $1$ is almost never written down, but it is there, and remembering it is what lets you handle $b \cdot b^8$.
• The exponent is a count, not a multiplier. $a^5$ is not $5a$. At $a = 3$ the first is $243$ and the second is $15$.

Multiplying powers of the same base

Now multiply two powers of $a$ and do nothing clever — just write out both rows of factors and count them.

The figure writes $a^3 \cdot a^2$ as a single row of $a$s. The blue brace covers 3 factors, the red brace covers 2 factors, and the black brace over the top covers 5 factors in all. So

$$a^3 \cdot a^2 = a^5$$

Nothing was assumed. The exponents added because the only thing that happened was counting how many copies of the base were standing in the row.

The table underneath runs the same count on four more products. $2^3 \cdot 2^2$ becomes $(2 \cdot 2 \cdot 2)(2 \cdot 2)$, a count of $3 + 2 = 5$, so $2^5$ — and the numeric check at the bottom of the figure confirms it: $2^3 \cdot 2^2 = 8 \cdot 4 = 32$, and $2^5 = 32$. The row $x^4 \cdot x^1$ shows the invisible exponent doing its job: $(x \cdot x \cdot x \cdot x)(x)$ is $4 + 1 = 5$ factors, so $x^5$. The shaded row does $a^2 \cdot a^6$, a count of $2 + 6 = 8$, giving $a^8$. The last row does the general case: $m$ factors, then $n$ factors, for a total of $m + n$.

The product law. For any base $a$ and any exponents $m$ and $n$, $$a^m \cdot a^n = a^{m+n}$$

The bases must match

The law counts copies of one base. If the bases differ, there is nothing to count together.

$$x^3 \cdot y^4 \ \text{stays} \ x^3y^4$$

There is no single power that this equals, because the expression is three copies of $x$ and four copies of $y$, and those are different things being multiplied. Writing $x^3 \cdot y^4 = (xy)^7$ claims seven copies of $xy$, which would be seven $x$s and seven $y$s.

When several bases appear at once, sort them and handle each separately. That is what makes the law useful on multivariable expressions, which is what A.EO.3b asks for:

$$x^3y^2 \cdot x^4y^5 = (x^3 \cdot x^4)(y^2 \cdot y^5) = x^7y^7$$

Coefficients are multiplied, not added

A coefficient is a number multiplying a power, and it plays no part in the counting of factors. It is simply multiplied.

$$3x^2 \cdot 5x^6 = (3 \cdot 5)(x^2 \cdot x^6) = 15x^8$$

The coefficients multiply because $3 \cdot 5$ really is $15$; the exponents add because $2$ copies of $x$ followed by $6$ copies of $x$ really is $8$ copies. Two different operations, on two different parts of the term, both coming straight from what the notation means.

Here is the procedure in full.

1. Match the bases. Group the powers that share a base; leave the rest alone.
2. Multiply the coefficients, signs included.
3. Add the exponents on each matched base, reading a bare variable as exponent $1$.
4. Write each base once, in a consistent order — alphabetical is the convention in this volume.

Where the minus sign lives

This is the classic error of the chapter, and it has nothing to do with the laws. It is a question about what the base is.

The parentheses decide what the base is. Read the table row by row:

Read $-a^n$ as "the opposite of $a^n$." That single rewording settles every case, because it puts the exponent where it actually is and applies the minus sign afterward.

Notice the trap the figure names explicitly: $(-2)^3$ and $-2^3$ both equal $-8$, so they agree only because the exponent is odd. At an even exponent they never agree — $(-2)^4 = 16$ while $-2^4 = -16$. A student who checks the rule once with an odd exponent will conclude the parentheses do not matter, and will be wrong every other time.

The same care applies with variables. In $3x^2$ only the $x$ is squared; in $-c^5 \cdot -8c^4$ the two minus signs multiply to a plus, giving $8c^9$.

Worked examples

Example 1 — One base

Write $m^5 \cdot m^7$ as a single power.

The bases match, so add the counts: $5 + 7 = 12$.

Answer: $m^{12}$

Example 2 — An invisible exponent

Write $b \cdot b^8$ as a single power.

Read $b$ as $b^1$. Then $1 + 8 = 9$.

Answer: $b^9$

Example 3 — A coefficient and a sign

Simplify $-4a^3 \cdot 6a^5$.

Multiply the coefficients: $-4 \cdot 6 = -24$. Add the exponents: $3 + 5 = 8$.

Answer: $-24a^8$

Example 4 — Two variables

Simplify $2m^2n \cdot 5m^6n^3$.

Coefficients: $2 \cdot 5 = 10$. Base $m$: $2 + 6 = 8$. Base $n$: $1 + 3 = 4$.

Answer: $10m^8n^4$

Example 5 — Which symbol has the exponent

Evaluate $(-2)^5$ and $-2^5$.

In the first, the base is $-2$: $(-2)(-2)(-2)(-2)(-2) = -32$. In the second, the base is $2$ and the minus sign is applied afterward: $-(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2) = -32$.

Answer: Both equal $-32$ — but only because the exponent $5$ is odd. At an even exponent they differ: $(-2)^6 = 64$ while $-2^6 = -64$.

Guided practice

1. Use the product-law figure. Write $a^3 \cdot a^2$ as a row of factors, say what each of the three braces counts, and give the single power.
2. In that same figure, read the shaded row. What are the factors of $a^2 \cdot a^6$ written out, what is the count, and what single power results?
3. Using the figure, explain in one or two sentences why the exponents add. Your explanation should mention counting and should not use the word "rule."
4. Write $2^3 \cdot 2^4$ as a single power, then evaluate both the original and your answer to confirm they agree.
5. Write $m^5 \cdot m^7$ as a single power.
6. Use the figure comparing negative bases. Give the value of $(-3)^2$ and the value of $-3^2$, and say in one sentence what makes them different.

Independent practice

7. Write each as a single power. a) $x^6 \cdot x^3$ b) $7^2 \cdot 7^5$ c) $b \cdot b^8$ d) $y^4 \cdot y^4$
8. Simplify. a) $3x^2 \cdot 5x^6$ b) $-4a^3 \cdot 6a^5$ c) $2p^7 \cdot p^2 \cdot 3p$ d) $(-c^5)(-8c^4)$
9. Simplify each multivariable product. a) $x^3y^2 \cdot x^4y^5$ b) $2m^2n \cdot 5m^6n^3$ c) $-3a^4b^2 \cdot a^2b^7$ d) $4r^2s^3t \cdot 3rs^2t^4$
10. Reasoning. Explain why $x^3 \cdot y^4$ cannot be written as a single power. Say what $(xy)^7$ would mean, and why that is a different expression.
11. Evaluate all four: $(-2)^5$, $-2^5$, $(-2)^6$, $-2^6$. Then say what the four values together show about when the parentheses matter.
12. Application. A backup archive holds $2^5$ files, and each file is exactly $2^{12}$ bytes. Write the total number of bytes as a single power of $2$, then give it as an ordinary number, and name the law you used.
13. Reasoning. Explain why $a^1 = a$, and then explain why the product law would fail on $b \cdot b^8$ if you refused to read $b$ as $b^1$.
14. Error analysis. A student writes $x^3 \cdot x^4 = x^{12}$. Identify the error, give the correct answer, and show the count of factors that settles it.
15. Error analysis. A student writes $-5^2 = 25$. Identify the error, give the correct value, and write the expression whose value really is $25$.
16. Find each missing exponent. a) $x^4 \cdot x^{\underline{\hspace{1cm}}} = x^{11}$ b) $3a^{\underline{\hspace{1cm}}} \cdot 4a^5 = 12a^9$ c) $m^2n^{\underline{\hspace{1cm}}} \cdot m^5n^3 = m^7n^8$

Exit ticket 10.1

17. Write $y^7 \cdot y^6$ as a single power.
18. Simplify $-2x^3y \cdot 7x^5y^4$.
19. Evaluate $(-4)^2$ and $-4^2$, and say which symbol carries the exponent in each.
20. Explain why the product law requires the two bases to match. Use $x^3 \cdot y^4$ as your example.

Lesson 10.2 — The Quotient Law

Dividing powers of the same base

Multiplication built a row of factors; division takes them away in pairs. Again, write everything out and look.

The figure writes $\dfrac{x^5}{x^2}$ with all seven factors visible: five $x$s on top, two on the bottom. Each of the bottom $x$s is struck through together with an $x$ above it, because a factor divided by itself is $1$. Two pairs are matched off that way, and the brace labels what survives: 3 factors left over. So

$$\frac{x^5}{x^2} = x^3$$

The count is $5 - 2 = 3$. Subtracting exponents is nothing more than bookkeeping for canceling pairs — the figure's own note says it: each pair matched off is a factor divided by itself, which is $1$.

The table runs the same argument three more ways. For $\dfrac{2^5}{2^2}$, two pairs cancel and three $2$s remain, so $5 - 2 = 3$ and the answer is $2^3$; the numeric check at the bottom of the figure confirms it, since $\dfrac{32}{4} = 8$ and $2^3 = 8$. For $\dfrac{y^7}{y^4}$, four pairs cancel and three $y$s remain, so $7 - 4 = 3$ and the answer is $y^3$. The last row states the general case: $n$ pairs cancel and $m - n$ factors remain.

The quotient law. For any base $a \neq 0$ and any exponents $m$ and $n$, $$\frac{a^m}{a^n} = a^{m-n}$$

Why $a \neq 0$

The restriction is not decoration. If $a = 0$, the denominator $a^n$ is $0$, and $\dfrac{0}{0}$ names no number at all. So the law is a statement about every base except zero, and every consequence of the law — including the two derivations of the next lesson — inherits that same exception. State it when you state the law; it will matter in Lesson 10.3.

Coefficients divide

The coefficients of a ratio of monomials are divided, and the result is reported as a reduced fraction or a whole number, whichever it turns out to be.

$$\frac{20x^8}{4x^3} = \frac{20}{4} \cdot \frac{x^8}{x^3} = 5x^5$$

$$\frac{15m^9}{25m^4} = \frac{15}{25} \cdot \frac{m^9}{m^4} = \frac{3m^5}{5}$$

The second one is worth pausing on. The coefficient fraction $\tfrac{15}{25}$ reduces to $\tfrac35$, and $\tfrac35$ is the honest answer; there is no reason for it to come out whole. Nothing about the exponents changed because of it.

The full procedure for a ratio of monomials:

1. Divide the coefficients, signs included, and reduce the fraction.
2. Subtract exponents base by base, top exponent minus bottom exponent, in that order.
3. Handle each base separately. A base that appears only on top, or only on the bottom, stays where it is.
4. Record the restriction. Every variable that appears in the denominator is nonzero.

Several bases at once

This is exactly what A.EO.3b means by a ratio of monomial expressions, and the work is the same work done once per base.

$$\frac{12a^6b^4}{4a^2b} = \frac{12}{4} \cdot \frac{a^6}{a^2} \cdot \frac{b^4}{b^1} = 3a^4b^3$$

Coefficients: $12 \div 4 = 3$. Base $a$: $6 - 2 = 4$. Base $b$: $4 - 1 = 3$, reading the bare $b$ as $b^1$. Three independent subtractions, one for each base, and a division for the numbers out front.

Keep the order of the subtraction straight. It is always numerator exponent minus denominator exponent. Reversing it on one base out of three is the most common way an otherwise correct simplification goes wrong.

What happens when the exponents are equal, or upside down

So far every example had a larger exponent on top, so the leftover count was positive. Two other cases exist, and it is worth naming them now even though the next lesson is where they get settled.

• Equal exponents. In $\dfrac{a^3}{a^3}$ every factor pairs off, so nothing is left over and the count is $3 - 3 = 0$.
• A larger exponent on the bottom. In $\dfrac{a^2}{a^5}$ the two top factors pair off, and three factors are left over on the bottom, so the count is $2 - 5 = -3$.

Do not reach for a rule you have not derived. Just notice that the counting still works perfectly well, and that it is producing exponents of $0$ and of $-3$. Lesson 10.3 is about what those two exponents have to mean.

Worked examples

Example 1 — One base

Write $\dfrac{m^9}{m^4}$ as a single power.

Four pairs cancel, five factors remain: $9 - 4 = 5$.

Answer: $m^5$, for $m \neq 0$

Example 2 — A coefficient

Simplify $\dfrac{12a^7}{3a^2}$.

Divide the coefficients: $12 \div 3 = 4$. Subtract the exponents: $7 - 2 = 5$.

Answer: $4a^5$

Example 3 — A negative coefficient

Simplify $\dfrac{-18a^6}{6a^2}$.

Coefficients: $-18 \div 6 = -3$. Exponents: $6 - 2 = 4$.

Answer: $-3a^4$

Example 4 — Two variables

Simplify $\dfrac{x^7y^5}{x^3y^2}$.

Base $x$: $7 - 3 = 4$. Base $y$: $5 - 2 = 3$.

Answer: $x^4y^3$

Example 5 — A coefficient that stays a fraction

Simplify $\dfrac{15m^9}{25m^4}$.

$\dfrac{15}{25}$ reduces to $\dfrac35$, and $9 - 4 = 5$.

Answer: $\dfrac{3m^5}{5}$

Guided practice

21. Use the quotient-law figure. Write $\dfrac{x^5}{x^2}$ with all its factors shown, say how many pairs are struck through, say how many factors are left over, and give the single power.
22. In that same figure, read the shaded row. For $\dfrac{y^7}{y^4}$, how many pairs cancel, how many $y$s remain, and what single power results?
23. Use the numeric check printed in that figure. Evaluate $\dfrac{2^5}{2^2}$ as a number, evaluate $2^3$, and say what the agreement confirms.
24. Write $\dfrac{m^9}{m^4}$ as a single power.
25. Simplify $\dfrac{12a^7}{3a^2}$.
26. Explain why the quotient law must say $a \neq 0$. Say exactly what goes wrong at $a = 0$.

Independent practice

27. Write each as a single power. a) $\dfrac{x^{10}}{x^4}$ b) $\dfrac{5^8}{5^3}$ c) $\dfrac{n^6}{n}$ d) $\dfrac{c^{12}}{c^5}$
28. Simplify. a) $\dfrac{20x^8}{4x^3}$ b) $\dfrac{-18a^6}{6a^2}$ c) $\dfrac{15m^9}{25m^4}$ d) $\dfrac{-24p^7}{-8p^3}$
29. Simplify each ratio of monomials. a) $\dfrac{x^7y^5}{x^3y^2}$ b) $\dfrac{12a^6b^4}{4a^2b}$ c) $\dfrac{-30m^8n^6}{5m^3n^2}$ d) $\dfrac{r^9s^4t^2}{r^4s^2t}$
30. Application. A stadium seats about $3 \times 10^4$ people and a school auditorium seats about $3 \times 10^2$. Write the ratio of the two capacities as a power of ten, and say in one sentence how many times as many people the stadium holds.
31. Application. A drive holds $2^{40}$ bytes and each backup image is $2^{28}$ bytes. Write the number of images that fit as a power of $2$, then give it as an ordinary number.
32. Reasoning. Explain why subtracting exponents does the same job as canceling matched pairs of factors. Use $\dfrac{x^5}{x^2}$ and say what each canceled pair is worth.
33. Error analysis. A student writes $\dfrac{x^8}{x^2} = x^4$. Identify what the student did to the exponents, give the correct answer, and check it by counting leftover factors.
34. Error analysis. A student writes $\dfrac{12a^6}{4a^2} = 8a^4$. Identify the error, and say which operation belongs to the coefficients and which belongs to the exponents.
35. Find each missing exponent. a) $\dfrac{x^{\underline{\hspace{1cm}}}}{x^3} = x^9$ b) $\dfrac{20m^7}{5m^{\underline{\hspace{1cm}}}} = 4m^3$ c) $\dfrac{a^8b^{\underline{\hspace{1cm}}}}{a^3b^4} = a^5b^2$
36. Reasoning. In $\dfrac{a^m}{a^n}$ with $m > n$, the leftover count $m - n$ is positive. Describe, in terms of leftover factors only, what happens when $m = n$ and what happens when $m < n$. Do not use any law you have not yet derived.

Exit ticket 10.2

37. Write $\dfrac{x^{11}}{x^6}$ as a single power.
38. Simplify $\dfrac{-28a^9b^5}{7a^4b^2}$.
39. Write $\dfrac{2^{12}}{2^7}$ as a single power of $2$, then give its value as a number.
40. Explain why every statement of the quotient law carries the restriction $a \neq 0$.

Lesson 10.3 — Zero and Negative Exponents

This is the heart of the chapter. The two facts in it — $a^0 = 1$ and $a^{-n} = \tfrac{1}{a^n}$ — are the two that students most often file away as arbitrary conventions someone decided on. They are not. They are forced. Once the product and quotient laws are in place, no other values are available, and a student who reads one quotient two ways cannot escape the conclusion.

The pattern that runs down to zero

Start with powers of $2$ and walk down the exponents one step at a time.

The left-hand table is the pattern. Read the right-hand column first, because it is the same instruction five times over:

Step down one exponent, divide by $2$. That is what dropping one factor of $2$ means, and it happens at every single step. The last row is shaded and the arrow beside it says forced, for a plain reason: if the rule for stepping down is "divide by $2$," then the value below $2^1 = 2$ is $2 \div 2 = 1$. Nobody chose that. Choosing anything else would mean the pattern broke on the very last step for no reason.

The same conclusion from one quotient read twice

The pattern makes $a^0 = 1$ overwhelmingly likely. The right-hand panel of the figure makes it unavoidable. Take a single expression, $\dfrac{a^3}{a^3}$, and read it two ways.

• By the quotient law: $\dfrac{a^3}{a^3} = a^{3-3} = a^0$.
• By canceling all three pairs: $\dfrac{a \cdot a \cdot a}{a \cdot a \cdot a} = 1$.

The same expression cannot equal two different things. Both readings are legitimate — the first uses a law derived in Lesson 10.2, the second uses only the fact that a factor over itself is $1$ — so the two answers must be the same number.

$$a^0 = 1 \qquad \text{for every } a \neq 0$$

That is a derivation, not a convention. It is also exactly why the restriction rides along: the argument divides by $a^3$, which is only allowed when $a \neq 0$.

Once you have it, $a^0 = 1$ is indifferent to what $a$ is. $7^0 = 1$. $(-5)^0 = 1$. $\left(\tfrac{2}{3}\right)^0 = 1$. And $(3x)^0 = 1$ for every $x \neq 0$, because whatever number $3x$ happens to be, it is being raised to the zero power.

$0^0$

The one base the derivation cannot reach is $0$, and it is worth being honest about why rather than quietly skipping it.

Two patterns collide there. Looking along the exponents, $a^0 = 1$ for every nonzero $a$, which suggests $0^0$ should be $1$. Looking along the bases, $0^n = 0$ for every positive $n$ — $0^3 = 0$, $0^2 = 0$, $0^1 = 0$ — which suggests $0^0$ should be $0$. There is no reading that satisfies both, and the derivation above cannot break the tie, because it divided by $a^3$ and $0^3 = 0$.

So $0^0$ is left undefined in this course. That is not a gap in the theory; it is the theory declining to invent an answer that no pattern forces.

A context whose zero row is a real moment

The zero exponent can feel like a technicality. Here is a situation where it is a plain fact about the world.

One cell divides in two every hour. After $h$ hours there are $2^h$ cells:

The boxed column is hour $0$ — the moment the observation begins. There is one cell then, because nothing has divided yet. And the power in that column is $2^0$. So $2^0 = 1$ is not a bookkeeping fiction; in this table it is a headcount. The zero exponent is what "before anything has happened yet" looks like when a quantity is written as a power.

(How this table behaves as a function — its domain, its graph, its rate of growth — is Chapter 17's subject. Here it is a column of arithmetic.)

Continuing the pattern below zero

Now do the obvious thing: keep walking down. The instruction has not changed.

The left-hand table is the same column as before, extended:

Look at the figure's own note: nothing new is decided at the line $2^0$. The three shaded rows below it are the next steps of a pattern that was already running. Divide $1$ by $2$ and you get $\tfrac12$, so $2^{-1} = \tfrac12$. Divide again and $2^{-2} = \tfrac14$. Again and $2^{-3} = \tfrac18$.

This is where the single most common misreading dies. A student expecting $2^{-1}$ to be $-2$ has to explain why the column suddenly stopped dividing by $2$ and started doing something else. There is no such explanation. A negative exponent is not a negative number. It is an instruction to take a reciprocal, and every value in the shaded rows is a positive number smaller than $1$.

The decimal column is there to be checked. Enter $2^{-3}$ on a calculator and it returns $0.125$. That takes one keystroke and it is worth doing once, because seeing $0.125$ rather than $-8$ or $-6$ settles the matter in a way an argument sometimes does not.

And the same two-readings argument

The pattern strongly suggests the rule; one quotient read twice forces it. The right-hand panel takes $\dfrac{a^2}{a^5}$.

• By the quotient law: $\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}$.
• By canceling the two pairs: $\dfrac{a \cdot a}{a \cdot a \cdot a \cdot a \cdot a} = \dfrac{1}{a^3}$, since two factors on top match two on the bottom and three factors are left over below the bar.

One expression, two correct readings, so the answers agree.

$$a^{-n} = \frac{1}{a^n} \qquad \text{for every } a \neq 0$$

Turn it over and the same statement reads $\dfrac{1}{a^{-n}} = a^n$: a negative exponent in a denominator moves up as a positive one. Both directions are the same fact, which is what "reciprocal" means.

Writing answers with positive exponents

The convention of this volume is that a final answer contains no negative exponent. Moving one is a matter of crossing the bar.

• A negative exponent in the numerator moves to the denominator and turns positive: $x^{-5} = \dfrac{1}{x^5}$.
• A negative exponent in the denominator moves to the numerator and turns positive: $\dfrac{1}{m^{-4}} = m^4$.
• Both at once: $\dfrac{a^{-3}}{b^{-2}} = \dfrac{b^2}{a^3}$.

One caution, because it accounts for most of the errors here: only the base with the negative exponent moves. A coefficient stays where it is.

$$3y^{-2} = \frac{3}{y^2} \qquad \text{not} \qquad \frac{1}{3y^2}$$

The exponent $-2$ sits on the $y$ alone, exactly as in Lesson 10.1. Check it at $y = 2$: the correct form gives $3 \cdot \tfrac14 = 0.75$, while the wrong one gives $\tfrac{1}{12}$, which is about $0.083$. Not close.

Worked examples

Example 1 — Zero exponents

Evaluate $9^0$, $(-5)^0$, and $(3x)^0$ for $x \neq 0$.

Every nonzero base raised to the zero power is $1$.

Answer: $1$, $1$, and $1$

Example 2 — A negative exponent as a fraction

Evaluate $4^{-2}$.

$4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16}$.

Answer: $\dfrac{1}{16}$, or $0.0625$ on a calculator

Example 3 — A negative base and a negative exponent

Evaluate $(-3)^{-3}$.

$(-3)^{-3} = \dfrac{1}{(-3)^3} = \dfrac{1}{-27} = -\dfrac{1}{27}$.

Answer: $-\dfrac{1}{27}$. The value is negative because the base is negative and the exponent is odd, not because the exponent is negative.

Example 4 — Combining laws

Simplify $x^{-3} \cdot x^7$, and write the answer with a positive exponent.

The product law adds exponents whatever their signs: $-3 + 7 = 4$.

Answer: $x^4$

Example 5 — Two bases, both moving

Simplify $\dfrac{a^3b^{-2}}{a^{-1}b^4}$ with positive exponents.

Base $a$: $3 - (-1) = 4$. Base $b$: $-2 - 4 = -6$. So the result is $a^4b^{-6}$, and the $b$ moves down.

Answer: $\dfrac{a^4}{b^6}$

Guided practice

41. Use the descending-pattern figure for the zero exponent. List the five powers in the table with their values, and give the division in the right-hand column that produces each value from the row above.
42. In that same figure, read the two-readings panel. What does the quotient law give for $\dfrac{a^3}{a^3}$? What does canceling all three pairs give? What must therefore be true, and why can it not be otherwise?
43. Use the cell-division figure. Which column is boxed, what number of cells does it record, and what power of $2$ sits in it? Say in one sentence why that column is a real moment in the story rather than a technicality.
44. Evaluate $7^0$, $(-5)^0$, and $(3x)^0$ for $x \neq 0$.
45. Use the figure that continues the pattern below zero. Give the values of $2^{-1}$, $2^{-2}$, and $2^{-3}$ as fractions and as decimals, and give the division that produces each from the row above.
46. In that same figure, read the two-readings panel. What does the quotient law give for $\dfrac{a^2}{a^5}$? What does canceling the two pairs give? What law follows?
47. Explain why both $a^0 = 1$ and $a^{-n} = \tfrac{1}{a^n}$ carry the restriction $a \neq 0$, and say why $0^0$ is left undefined.

Independent practice

48. Evaluate each exactly. a) $9^0$ b) $4^{-2}$ c) $(-3)^{-3}$ d) $\left(\tfrac{2}{5}\right)^{-1}$
49. Give each as a decimal, then confirm it on a calculator and say what you entered. a) $2^{-3}$ b) $5^{-2}$ c) $10^{-4}$
50. Rewrite each with positive exponents only. a) $x^{-5}$ b) $3y^{-2}$ c) $\dfrac{1}{m^{-4}}$ d) $\dfrac{a^{-3}}{b^{-2}}$
51. Simplify, writing each answer with positive exponents. a) $x^{-3} \cdot x^7$ b) $\dfrac{m^2}{m^6}$ c) $a^{-4} \cdot a^{-3}$ d) $\dfrac{y^{-2}}{y^5}$
52. Simplify each multivariable expression with positive exponents. a) $x^{-2}y^5 \cdot x^6y^{-3}$ b) $\dfrac{a^3b^{-2}}{a^{-1}b^4}$ c) $\dfrac{6m^{-3}n^4}{2m^2n^{-1}}$ d) $4p^0q^{-2} \cdot 5p^3q^2$
53. Reasoning. Explain why $3x^{-2}$ is $\dfrac{3}{x^2}$ and not $\dfrac{1}{3x^2}$. Then evaluate both at $x = 2$ and report the two values, to show they are not the same expression.
54. Application. Use the cell-division figure. Explain why the hour-$0$ column records $1$ cell, give the number of cells after $6$ hours, and give the number after $10$ hours as a power of $2$ and as an ordinary number.
55. Application. A thumbnail image takes up $2^{-3}$ megabytes of storage. Write that size as a fraction and as a decimal, and say how many thumbnails fit in one megabyte.
56. Reasoning. Using the divide-by-$2$ column of the figure that continues the pattern below zero, explain why $2^{-1}$ has to be $\tfrac12$ and cannot be $-2$. Your explanation should say what would have to happen to the pattern for $-2$ to be right.
57. Reasoning. Explain why $a^0 = 1$ is forced rather than chosen. Use the two readings of $\dfrac{a^3}{a^3}$, and say where the restriction $a \neq 0$ enters the argument.
58. Error analysis. A student says $5^{-2} = -25$. Identify the two separate mistakes in that answer, give the correct value, and say what expression really does equal $-25$.
59. Error analysis. A student says $x^0 = 0$ "because the exponent is zero." Identify the error, give the correct value, and name the quotient that forces it.
60. Reasoning. Explain why $0^0$ is left undefined. Name the two patterns that disagree there, and say why the derivation of $a^0 = 1$ cannot settle the case $a = 0$.

Exit ticket 10.3

61. Evaluate $12^0$, $3^{-4}$, and $(-2)^{-4}$.
62. Rewrite $\dfrac{5x^{-3}}{y^{-2}}$ with positive exponents.
63. Simplify $\dfrac{a^2b^{-5}}{a^{-3}b^{-1}}$ with positive exponents.
64. Complete the column: $2^3 = 8$, $2^2 = 4$, $2^1 = 2$, $2^0 = \underline{\hspace{2cm}}$, $2^{-1} = \underline{\hspace{2cm}}$. Then state in one sentence the single rule that produces every entry from the one above it.
65. Explain in two sentences why $a^0 = 1$ is forced by the quotient law rather than chosen as a convention.

Lesson 10.4 — Powers of Bases

A.EO.3a names three things to derive from patterns: products, quotients, and powers of bases. The first two are done. This lesson does the third, and there are three of them: a power of a power, a power of a product, and a power of a quotient. All three come from counting, and none of them is new mathematics.

A power of a power counts groups

The figure writes $(a^2)^3$ out in full. The outer exponent $3$ says "three copies of the thing inside," and the thing inside is $a^2$, which is two factors. So:

$$(a^2)^3 = (a \cdot a)(a \cdot a)(a \cdot a) = a^6$$

The brace labels the count exactly: 3 groups of 2 factors, so 6 factors. The outer exponent counts groups; the inner exponent counts factors inside a group; and the total number of factors is groups times factors per group. That is multiplication, not addition, because it is the same arithmetic as counting three rows of two chairs.

The table repeats the count. $(2^3)^2$ is $2$ groups of $3$, so $3 \cdot 2 = 6$ factors and the answer is $2^6$ — confirmed by the numeric check in the figure, since $(2^3)^2 = 8^2 = 64$ and $2^6 = 64$. $(x^4)^3$ is $3$ groups of $4$, so $4 \cdot 3 = 12$ and the answer is $x^{12}$. The general row: $(a^m)^n$ is $n$ groups of $m$.

Power of a power. $(a^m)^n = a^{mn}$

This is the law most often confused with the product law, so it is worth holding the two side by side.

$$a^2 \cdot a^3 = a^5 \qquad\qquad (a^2)^3 = a^6$$

In the first, two rows of factors are laid end to end, so the counts add. In the second, one row of two is repeated three times, so the counts multiply. Different pictures, different arithmetic, and writing the factors out tells you which one you are looking at every time.

A power of a product, and a power of a quotient

Both panels of the figure run the same three moves: write the repeated factors out, regroup them, read off the result.

Left panel — a power of a product.

$$(2x)^3 = (2x)(2x)(2x) = (2 \cdot 2 \cdot 2)(x \cdot x \cdot x) = 8x^3$$

The regrouping in the middle is the whole derivation, and it is allowed because multiplication can be reordered. Once the $2$s are gathered and the $x$s are gathered, each group is a power in its own right. So each factor inside the parentheses takes the exponent separately:

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

The figure checks it at $x = 5$: $(2 \cdot 5)^3 = 10^3 = 1000$, and $8 \cdot 5^3 = 8 \cdot 125 = 1000$.

Right panel — a power of a quotient.

$$\left(\frac{x}{y}\right)^4 = \frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y} = \frac{x \cdot x \cdot x \cdot x}{y \cdot y \cdot y \cdot y} = \frac{x^4}{y^4}$$

Multiplying fractions multiplies the numerators and multiplies the denominators, so both take the exponent:

$$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}, \qquad b \neq 0$$

The figure checks that one at $x = 3$, $y = 2$: $\left(\tfrac{3}{2}\right)^4 = \tfrac{81}{16}$, and $\tfrac{3^4}{2^4} = \tfrac{81}{16}$. The restriction $b \neq 0$ is there for the reason it is always there — a denominator of zero names nothing.

Everything inside, including the coefficient

The most frequent error in this lesson is leaving a coefficient behind. $(ab)^n = a^nb^n$ applies to numbers as much as to variables, because a coefficient is a factor like any other.

$$(3x)^4 = 3^4 x^4 = 81x^4 \qquad \text{not} \qquad 3x^4$$

Check it at $x = 1$: the correct form gives $81$ and the wrong one gives $3$. When several factors sit inside, every one of them takes the exponent, and the power-of-a-power law handles those that are already powers:

$$(5m^2n)^3 = 5^3 (m^2)^3 n^3 = 125m^6n^3$$

Signs follow the same rule they followed in Lesson 10.1 — the base is whatever is inside the parentheses:

$$(-4p^3q^2)^2 = (-4)^2 (p^3)^2 (q^2)^2 = 16p^6q^4$$

$$(-2a)^5 = (-2)^5 a^5 = -32a^5$$

The first is positive because the exponent is even; the second is negative because it is odd.

Here is the procedure.

1. Distribute the outer exponent to every factor inside, the coefficient included.
2. Multiply exponents wherever a power is being raised to a power.
3. Raise the coefficient, and get its sign from whether the exponent is even or odd.
4. Simplify the coefficient to a number, so $3^4$ is reported as $81$.

Negative outer exponents

The outer exponent may itself be negative — A.EO.3b allows any integer — and nothing changes except that the answer needs its exponents made positive at the end.

$$(x^4)^{-3} = x^{-12} = \frac{1}{x^{12}}$$

$$(2m)^{-3} = 2^{-3}m^{-3} = \frac{1}{8m^3}$$

Notice that the coefficient took the negative exponent too, so $2^{-3} = \tfrac18$ ended up in the denominator. And a negative exponent on a fraction simply flips it, which is worth seeing once:

$$\left(\frac{a}{b}\right)^{-2} = \frac{a^{-2}}{b^{-2}} = \frac{b^2}{a^2}$$

One thing these laws do not do

$(ab)^n = a^nb^n$ is a statement about a product inside the parentheses. It says nothing about a sum, and the corresponding claim about sums is false.

$$(a + b)^n \neq a^n + b^n$$

At $a = 3$, $b = 4$, $n = 2$: the left side is $(3+4)^2 = 7^2 = 49$, and the right side is $3^2 + 4^2 = 9 + 16 = 25$. One counterexample is enough to retire a claim permanently. The reason is visible in the derivation: the regrouping step reordered factors, and there are no factors to reorder in $a + b$. Chapter 12 works out what $(a+b)^2$ actually is.

Worked examples

Example 1 — A power of a power

Simplify $(m^5)^4$.

Four groups of five factors: $5 \cdot 4 = 20$.

Answer: $m^{20}$

Example 2 — A coefficient inside

Simplify $(3x)^4$.

Every factor takes the exponent: $3^4 = 81$ and $x^4$.

Answer: $81x^4$

Example 3 — Several factors, one of them negative

Simplify $(-4p^3q^2)^2$.

$(-4)^2 = 16$; $(p^3)^2 = p^6$; $(q^2)^2 = q^4$.

Answer: $16p^6q^4$

Example 4 — A power of a quotient

Simplify $\left(\dfrac{2a}{b}\right)^4$.

Numerator: $(2a)^4 = 16a^4$. Denominator: $b^4$.

Answer: $\dfrac{16a^4}{b^4}$, for $b \neq 0$

Example 5 — A negative outer exponent

Simplify $(3x^{-2})^{-1}$ with positive exponents.

Every factor takes the $-1$: $3^{-1} = \tfrac13$, and $(x^{-2})^{-1} = x^2$.

Answer: $\dfrac{x^2}{3}$

Guided practice

66. Use the power-of-a-power figure. Write $(a^2)^3$ as three groups of factors, say what the brace counts, and give the single power.
67. In that same figure, read the shaded row. For $(x^4)^3$, how many groups are there, how many factors are in each, and what single power results?
68. Use the numeric check printed in that figure. Evaluate $(2^3)^2$ step by step, evaluate $2^6$, and say what the agreement confirms.
69. Use the left panel of the power-of-a-product figure. Write the three lines of the derivation of $(2x)^3$, then state the law the panel boxes.
70. Use the right panel of that figure. Write the derivation of $\left(\dfrac{x}{y}\right)^4$, state the law it boxes, and say why that law carries the restriction $b \neq 0$.
71. Reasoning. Explain why $(a^2)^3$ multiplies the exponents while $a^2 \cdot a^3$ adds them. Give both answers, and describe the different picture of factors behind each.

Independent practice

72. Simplify. a) $(m^5)^4$ b) $(2^4)^3$ c) $(y^7)^2$ d) $\big((a^2)^3\big)^2$
73. Simplify. a) $(3x)^4$ b) $(-2a)^5$ c) $(5m^2n)^3$ d) $(-4p^3q^2)^2$
74. Simplify. a) $\left(\dfrac{x}{3}\right)^3$ b) $\left(\dfrac{2a}{b}\right)^4$ c) $\left(\dfrac{m^3}{n^2}\right)^5$ d) $\left(\dfrac{3x^2}{y^4}\right)^2$
75. Simplify, writing each answer with positive exponents. a) $(x^4)^{-3}$ b) $(2m)^{-3}$ c) $\left(\dfrac{a}{b}\right)^{-2}$ d) $(3x^{-2})^{-1}$
76. Application. A storage cube has edge length $2x^3$ centimeters. Write its volume as a simplified monomial, and name the two laws you used.
77. Application. One kilometer is $10^3$ meters. Write the number of square meters in one square kilometer as a power of ten, then as an ordinary number, and name the law you used.
78. Reasoning. Explain why $(ab)^n = a^nb^n$ is true but $(a + b)^n = a^n + b^n$ is false. Give the counterexample at $a = 3$, $b = 4$, $n = 2$ with both values, and say which step of the derivation fails for a sum.
79. Error analysis. A student writes $(3x)^2 = 3x^2$. Identify the error, give the correct answer, and evaluate both at $x = 1$ to show they differ.
80. Error analysis. A student writes $(x^3)^4 = x^7$. Identify which law the student used, give the correct answer, and say what the correct count of factors is.
81. Find each missing exponent. a) $\left(x^{\underline{\hspace{1cm}}}\right)^5 = x^{20}$ b) $\left(2a^3\right)^{\underline{\hspace{1cm}}} = 8a^9$ c) $\left(\dfrac{m^4}{n}\right)^3 = \dfrac{m^{\underline{\hspace{1cm}}}}{n^{\underline{\hspace{1cm}}}}$