Appendix A — Answer Key, Chapter 10: The Laws of Exponents

SOL A.EO.3 (a, b) · Covers textbook Chapter 10 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 126 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below. In $a^n$, $a$ is the base, $n$ is the exponent, and the whole thing is a power. The exponent reaches only the symbol it sits on, so $-3^2 = -9$ while $(-3)^2 = 9$. A bare variable is read as exponent $1$. Final answers use positive exponents, so $\tfrac{x^3}{x^7}$ is reported as $\tfrac{1}{x^4}$ rather than $x^{-4}$. Every law derived from a quotient carries $a \neq 0$; the restriction is stated where it is the point of the item and assumed silently elsewhere. $0^0$ is undefined. All exponents anywhere in this chapter are integers — no answer uses a fractional exponent or a radical.

The laws, for reference:

$$a^m \cdot a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad a^0 = 1 \qquad a^{-n} = \frac{1}{a^n}$$

$$(a^m)^n = a^{mn} \qquad (ab)^n = a^nb^n \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$$

The figures used repeatedly in the chapter:

• Figure 1 derives the product law by counting a row of factors: $a^3 \cdot a^2$ as five $a$s, braced $3$, $2$, and $5$
• Figure 2 derives the quotient law by striking off matched pairs: $\tfrac{x^5}{x^2}$ with two pairs canceled and three factors left over
• Figure 3 forces $a^0 = 1$ — a table stepping $2^4 = 16$ down to $2^0 = 1$ by repeated division by $2$, beside $\tfrac{a^3}{a^3}$ read two ways
• Figure 4 forces $a^{-n} = \tfrac{1}{a^n}$ — the same column continued to $2^{-3} = \tfrac18 = 0.125$, beside $\tfrac{a^2}{a^5}$ read two ways
• Figure 5 derives $(a^m)^n = a^{mn}$ by counting groups: $(a^2)^3$ as three groups of two factors
• Figure 6 derives $(ab)^n = a^nb^n$ from $(2x)^3$ and $\left(\tfrac{a}{b}\right)^n = \tfrac{a^n}{b^n}$ from $\left(\tfrac{x}{y}\right)^4$
• Figure 7 separates $(-a)^n$ from $-a^n$ across six expressions
• Figure 8 is the powers-of-ten scale from $10^{-3}$ to $10^7$, in meters
• Figure 9 is the cell-division table, $2^h$ cells after $h$ hours, hours $0$ through $6$
• Figure 10 is the pair of blank pattern tables for student work

Lesson 10.1 — Powers and the Product Law

Guided practice

1. The row is $a \cdot a \cdot a \cdot a \cdot a$. The blue brace counts the 3 factors that came from $a^3$; the red brace counts the 2 factors that came from $a^2$; the black brace over the top counts 5 factors in all. The single power is $a^5$.
2. $a^2 \cdot a^6 = (a \cdot a)(a \cdot a \cdot a \cdot a \cdot a \cdot a)$. The count is $2 + 6 = 8$, so the single power is $a^8$.
3. Because the only thing happening is counting. $a^3$ is a row of three copies of $a$ and $a^2$ is a row of two copies; setting one row after the other makes one row of five copies, and $3 + 2 = 5$ is how many copies are standing there. The exponent is a tally of factors, so putting two tallies together adds them.
4. $2^3 \cdot 2^4 = 2^7$. Check: $2^3 \cdot 2^4 = 8 \cdot 16 = 128$, and $2^7 = 128$. They agree.
5. $m^5 \cdot m^7 = m^{12}$, since $5 + 7 = 12$.
6. $(-3)^2 = (-3)(-3) = 9$, and $-3^2 = -(3 \cdot 3) = -9$. The parentheses decide the base: in the first, the whole number $-3$ is squared; in the second, only the $3$ is squared and the minus sign is applied afterward.

Independent practice

7. a) $x^6 \cdot x^3 = x^9$ ($6 + 3$) b) $7^2 \cdot 7^5 = 7^7$ ($2 + 5$) c) $b \cdot b^8 = b^9$, reading $b$ as $b^1$, so $1 + 8 = 9$ d) $y^4 \cdot y^4 = y^8$ ($4 + 4$) — the exponents add even when they are equal; this is not $y^{16}$
8. a) $3x^2 \cdot 5x^6 = (3 \cdot 5)(x^{2+6}) = 15x^8$ b) $-4a^3 \cdot 6a^5 = (-4 \cdot 6)(a^{3+5}) = -24a^8$ c) $2p^7 \cdot p^2 \cdot 3p = (2 \cdot 1 \cdot 3)(p^{7+2+1}) = 6p^{10}$ d) $(-c^5)(-8c^4) = (-1)(-8)(c^{5+4}) = 8c^9$ — two negative coefficients multiply to a positive one
9. a) $x^3y^2 \cdot x^4y^5 = x^{3+4}y^{2+5} = x^7y^7$ b) $2m^2n \cdot 5m^6n^3 = 10m^{2+6}n^{1+3} = 10m^8n^4$ c) $-3a^4b^2 \cdot a^2b^7 = -3a^{4+2}b^{2+7} = -3a^6b^9$ d) $4r^2s^3t \cdot 3rs^2t^4 = 12r^{2+1}s^{3+2}t^{1+4} = 12r^3s^5t^5$
10. There is nothing to count together. $x^3 \cdot y^4$ is three copies of $x$ multiplied by four copies of $y$, and those are different bases, so no single tally describes the whole product. It stays $x^3y^4$. Writing $(xy)^7$ would claim seven copies of the pair $xy$ — that is, seven $x$s and seven $y$s — which is a different expression entirely. At $x = 2$, $y = 1$: the original is $8$, and $(xy)^7 = 2^7 = 128$.
11. $(-2)^5 = -32$; $-2^5 = -32$; $(-2)^6 = 64$; $-2^6 = -64$. The first pair agrees and the second does not. The parentheses matter whenever the exponent is even: an even power of a negative base is positive, while $-a^n$ is negative whatever $n$ is. At an odd exponent the two forms happen to agree, which is exactly why checking the rule once with an odd exponent misleads.
12. Total bytes $= 2^5 \cdot 2^{12} = 2^{17}$, by the product law, since $5 + 12 = 17$. As an ordinary number, $2^{17} = 131{,}072$ bytes.
13. $a^1 = a$ because an exponent counts how many copies of the base are multiplied, and one copy of $a$ is just $a$. The product law needs it because $b$ carries an invisible exponent of $1$: the count in $b \cdot b^8$ is $1 + 8 = 9$. A student who treats the bare $b$ as having no exponent has nothing to add, and either leaves the answer as $b \cdot b^8$ or writes $b^8$, losing a factor.
14. The student multiplied the exponents instead of adding them. The product law adds, because the two rows of factors are laid end to end. Correct: $x^3 \cdot x^4 = x^7$. The count settles it — $(x \cdot x \cdot x)(x \cdot x \cdot x \cdot x)$ is seven $x$s, not twelve.
15. The student squared $-5$, but no parentheses were written, so the exponent sits on the $5$ alone. $-5^2 = -(5 \cdot 5) = -25$. The expression whose value is $25$ is $(-5)^2$.
16. a) $x^4 \cdot x^7 = x^{11}$, since $4 + 7 = 11$, so the missing exponent is $\mathbf{7}$ b) $3a^4 \cdot 4a^5 = 12a^9$, since $4 + 5 = 9$, so the missing exponent is $\mathbf{4}$ c) $m^2n^5 \cdot m^5n^3 = m^7n^8$, since $5 + 3 = 8$, so the missing exponent is $\mathbf{5}$

Exit ticket 10.1

17. $y^7 \cdot y^6 = y^{13}$
18. $-2x^3y \cdot 7x^5y^4 = (-2 \cdot 7)(x^{3+5})(y^{1+4}) = -14x^8y^5$
19. $(-4)^2 = (-4)(-4) = 16$; the parentheses make $-4$ the base. $-4^2 = -(4 \cdot 4) = -16$; here the base is $4$ and the minus sign is applied to the finished power.
20. The product law counts copies of one base, so there must be one base to count. In $x^3 \cdot y^4$ there are three copies of $x$ and four copies of $y$; adding $3 + 4$ would claim seven copies of something, and there is no single thing there are seven of. When bases differ, each is tallied separately and the expression stays a product: $x^3y^4$.

Lesson 10.2 — The Quotient Law

Guided practice

21. $\dfrac{x^5}{x^2} = \dfrac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x}$. Two pairs are struck through, one for each factor in the denominator. Three factors are left over on top, which the brace labels. The single power is $x^3$, and the count is $5 - 2 = 3$.
22. Four pairs cancel and three $y$s remain, so $7 - 4 = 3$ and the single power is $y^3$.
23. $\dfrac{2^5}{2^2} = \dfrac{32}{4} = 8$, and $2^3 = 8$. The agreement confirms that subtracting the exponents produced the right number, not merely a plausible-looking symbol — the law was derived by counting, and the arithmetic independently agrees with the count.
24. $\dfrac{m^9}{m^4} = m^5$, since four pairs cancel and five factors remain: $9 - 4 = 5$. (For $m \neq 0$.)
25. $\dfrac{12a^7}{3a^2} = \dfrac{12}{3} \cdot a^{7-2} = 4a^5$
26. If $a = 0$, then $a^n = 0$ and the expression $\dfrac{a^m}{a^n}$ has a denominator of zero, which names no number — there is no value $q$ with $0 \cdot q = a^m$ when $a^m$ is also $0$, and no value at all when it is not. So the law is a claim about every base except zero, and the restriction has to be stated rather than assumed.

Independent practice

27. a) $\dfrac{x^{10}}{x^4} = x^6$ ($10 - 4$) b) $\dfrac{5^8}{5^3} = 5^5$ ($8 - 3$) c) $\dfrac{n^6}{n} = n^5$, reading the denominator as $n^1$, so $6 - 1 = 5$ d) $\dfrac{c^{12}}{c^5} = c^7$ ($12 - 5$)
28. a) $\dfrac{20x^8}{4x^3} = \dfrac{20}{4} \cdot x^{8-3} = 5x^5$ b) $\dfrac{-18a^6}{6a^2} = \dfrac{-18}{6} \cdot a^{6-2} = -3a^4$ c) $\dfrac{15m^9}{25m^4} = \dfrac{15}{25} \cdot m^{9-4} = \dfrac{3m^5}{5}$ — the coefficient fraction reduces to $\tfrac35$ and stays a fraction; nothing requires it to come out whole d) $\dfrac{-24p^7}{-8p^3} = \dfrac{-24}{-8} \cdot p^{7-3} = 3p^4$ — two negatives divide to a positive
29. a) $\dfrac{x^7y^5}{x^3y^2} = x^{7-3}y^{5-2} = x^4y^3$ b) $\dfrac{12a^6b^4}{4a^2b} = 3a^{6-2}b^{4-1} = 3a^4b^3$, reading the lone $b$ as $b^1$ c) $\dfrac{-30m^8n^6}{5m^3n^2} = -6m^{8-3}n^{6-2} = -6m^5n^4$ d) $\dfrac{r^9s^4t^2}{r^4s^2t} = r^{9-4}s^{4-2}t^{2-1} = r^5s^2t$
30. $\dfrac{3 \times 10^4}{3 \times 10^2} = \dfrac{3}{3} \times 10^{4-2} = 10^2$. The stadium holds about $100$ times as many people as the auditorium — two orders of magnitude more.
31. $\dfrac{2^{40}}{2^{28}} = 2^{40-28} = 2^{12}$ images, which is $4096$ images.
32. Every canceled pair is one factor divided by itself, and a factor over itself is $1$ — so removing a pair changes nothing about the value while removing one factor from the top count and one from the bottom count. In $\dfrac{x^5}{x^2}$ there are two factors on the bottom, so exactly two pairs can be matched off, and $5 - 2 = 3$ factors are left. The subtraction is simply the record of how many pairings were possible; it is bookkeeping for the canceling, not a separate rule.
33. The student divided the exponents, $8 \div 2 = 4$, instead of subtracting them. Correct: $\dfrac{x^8}{x^2} = x^{8-2} = x^6$. Counting leftovers settles it: two of the eight top factors pair off with the two bottom factors, leaving six on top.
34. The student subtracted the coefficients, $12 - 4 = 8$, instead of dividing them. Coefficients are divided, because they are numbers being divided; exponents are subtracted, because they are counts of paired-off factors. Correct: $\dfrac{12a^6}{4a^2} = \dfrac{12}{4} \cdot a^{6-2} = 3a^4$.
35. a) $\dfrac{x^{12}}{x^3} = x^9$, since $12 - 3 = 9$, so the missing exponent is $\mathbf{12}$ b) $\dfrac{20m^7}{5m^4} = 4m^3$, since $7 - 4 = 3$, so the missing exponent is $\mathbf{4}$ c) $\dfrac{a^8b^6}{a^3b^4} = a^5b^2$, since $6 - 4 = 2$, so the missing exponent is $\mathbf{6}$
36. When $m = n$: every factor on top pairs off with one on the bottom, so nothing is left over at all. The count $m - n$ is $0$. When $m < n$: all $m$ top factors pair off, and there are $n - m$ factors left over on the bottom, below the bar. The count $m - n$ is negative, and its size says how many unmatched factors remain in the denominator. In both cases the counting still works perfectly; what it produces is an exponent of $0$ or a negative exponent, and Lesson 10.3 settles what those have to mean.

Exit ticket 10.2

37. $\dfrac{x^{11}}{x^6} = x^5$
38. $\dfrac{-28a^9b^5}{7a^4b^2} = \dfrac{-28}{7} \cdot a^{9-4}b^{5-2} = -4a^5b^3$
39. $\dfrac{2^{12}}{2^7} = 2^{12-7} = 2^5$, which is $32$. (Checking directly: $\tfrac{4096}{128} = 32$.)
40. Because the law has $a^n$ in a denominator, and a denominator of zero names no number. If $a = 0$ then $a^n = 0$ and the quotient is meaningless, so the derivation — matching factors off in pairs, each pair worth $1$ — never applies at $a = 0$. Every consequence of the law inherits the same exception, including $a^0 = 1$ and $a^{-n} = \tfrac{1}{a^n}$.

Lesson 10.3 — Zero and Negative Exponents

Guided practice

41. Power	Value	From the row above
    $2^4$	$16$	—
    $2^3$	$8$	$16 \div 2$
    $2^2$	$4$	$8 \div 2$
    $2^1$	$2$	$4 \div 2$
    $2^0$	$1$	$2 \div 2$

    Every entry in the right-hand column is the same instruction: step down one exponent, divide by $2$. That is what dropping one factor of $2$ means.

42. By the quotient law, $\dfrac{a^3}{a^3} = a^{3-3} = a^0$. By canceling all three pairs, $\dfrac{a \cdot a \cdot a}{a \cdot a \cdot a} = 1$. Therefore $a^0 = 1$, for every $a \neq 0$. It cannot be otherwise because the same expression cannot equal two different things: both readings are legitimate — one uses the quotient law, the other uses only that a factor over itself is $1$ — so the two answers are the same number. Nothing was chosen; the value was forced.

43. The hour-$0$ column is boxed. It records $1$ cell, and the power sitting in it is $2^0$. It is a real moment because hour $0$ is when the observation begins and nothing has divided yet — there genuinely is one cell — so $2^0 = 1$ is a headcount in this table rather than a bookkeeping convention.

44. $7^0 = 1$, $(-5)^0 = 1$, and $(3x)^0 = 1$ for every $x \neq 0$. The law does not care what the base is, only that it is not zero; whatever number $3x$ happens to be, it is being raised to the zero power.

45. Power	Value	As a decimal	From the row above
    $2^{-1}$	$\tfrac12$	$0.5$	$1 \div 2$
    $2^{-2}$	$\tfrac14$	$0.25$	$\tfrac12 \div 2$
    $2^{-3}$	$\tfrac18$	$0.125$	$\tfrac14 \div 2$

    The instruction is unchanged from the rows above $2^0$: divide by $2$.

46. By the quotient law, $\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}$. By canceling the two pairs, $\dfrac{a \cdot a}{a \cdot a \cdot a \cdot a \cdot a} = \dfrac{1}{a^3}$ — the two top factors match two of the bottom five, and three unmatched factors remain below the bar. So $a^{-n} = \dfrac{1}{a^n}$ for every $a \neq 0$.

47. Both are derived from a quotient. The argument for $a^0 = 1$ divides by $a^3$, and the argument for $a^{-n} = \tfrac{1}{a^n}$ divides by $a^5$; neither division is allowed when $a = 0$, so neither conclusion reaches that base. $0^0$ is left undefined because two patterns disagree there and nothing breaks the tie: reading along the exponents, $a^0 = 1$ for every nonzero $a$, which suggests $1$; reading along the bases, $0^n = 0$ for every positive $n$, which suggests $0$. Rather than invent an answer no pattern forces, this course declines to assign one.

Independent practice

48. a) $9^0 = 1$ b) $4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16}$ (which is $0.0625$) c) $(-3)^{-3} = \dfrac{1}{(-3)^3} = \dfrac{1}{-27} = -\dfrac{1}{27}$. The value is negative because the base is negative and the exponent is odd — not because the exponent is negative. d) $\left(\tfrac{2}{5}\right)^{-1} = \dfrac{1}{\ \tfrac{2}{5}\ } = \dfrac{5}{2}$. An exponent of $-1$ is exactly "take the reciprocal."

49. a) $2^{-3} = \tfrac18 = 0.125$. Entered as 2^(-3), or as $1 \div 8$. b) $5^{-2} = \tfrac{1}{25} = 0.04$. Entered as 5^(-2). c) $10^{-4} = \tfrac{1}{10000} = 0.0001$. Entered as 10^(-4).

    In all three cases the calculator returns a positive number smaller than $1$, which is the fastest possible refutation of reading a negative exponent as a negative number.

50. a) $x^{-5} = \dfrac{1}{x^5}$ b) $3y^{-2} = \dfrac{3}{y^2}$ — only the $y$ moves, because the exponent sits on the $y$ alone c) $\dfrac{1}{m^{-4}} = m^4$ — a negative exponent in a denominator moves up and turns positive d) $\dfrac{a^{-3}}{b^{-2}} = \dfrac{b^2}{a^3}$ — both cross the bar, in opposite directions

51. a) $x^{-3} \cdot x^7 = x^{-3+7} = x^4$ b) $\dfrac{m^2}{m^6} = m^{2-6} = m^{-4} = \dfrac{1}{m^4}$ c) $a^{-4} \cdot a^{-3} = a^{-4+(-3)} = a^{-7} = \dfrac{1}{a^7}$ d) $\dfrac{y^{-2}}{y^5} = y^{-2-5} = y^{-7} = \dfrac{1}{y^7}$

52. a) $x^{-2}y^5 \cdot x^6y^{-3} = x^{-2+6}y^{5+(-3)} = x^4y^2$ b) $\dfrac{a^3b^{-2}}{a^{-1}b^4} = a^{3-(-1)}b^{-2-4} = a^4b^{-6} = \dfrac{a^4}{b^6}$. Note $3 - (-1) = 4$: subtracting a negative exponent adds. c) $\dfrac{6m^{-3}n^4}{2m^2n^{-1}} = 3m^{-3-2}n^{4-(-1)} = 3m^{-5}n^5 = \dfrac{3n^5}{m^5}$ d) $4p^0q^{-2} \cdot 5p^3q^2 = 20 \cdot p^{0+3} \cdot q^{-2+2} = 20p^3q^0 = 20p^3$. Both special exponents appear: $p^0 = 1$ contributed nothing to the front, and $q^0 = 1$ removed $q$ entirely.

53. The exponent $-2$ sits on the $x$ alone, so only the $x$ moves across the bar; the coefficient $3$ is a separate factor and stays in the numerator. Hence $3x^{-2} = \dfrac{3}{x^2}$. At $x = 2$: the correct form gives $\dfrac{3}{4} = 0.75$, while $\dfrac{1}{3x^2} = \dfrac{1}{12} \approx 0.083$. The two are not the same expression, and one substitution shows it.

54. The hour-$0$ column records $1$ cell because hour $0$ is the moment the observation begins and no division has happened yet — there is still the one original cell. After $6$ hours there are $2^6 = 64$ cells, as the last column of the figure shows. After $10$ hours there are $2^{10}$ cells, which is $1024$.

55. $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$ megabyte, which is $0.125$ megabyte. Since $\dfrac{1}{\ 1/8\ } = 8$, eight thumbnails fit in one megabyte. (Equivalently, $\tfrac{2^0}{2^{-3}} = 2^3 = 8$.)

56. The right-hand column of that table says the same thing on every row: divide by $2$. The row above $2^{-1}$ is $2^0 = 1$, so the value at $2^{-1}$ is $1 \div 2 = \tfrac12$. For $2^{-1}$ to be $-2$, the pattern would have to abandon dividing by $2$ at exactly that row and start doing something else — negating and doubling — with no reason given, and then presumably do something different again at the next row. Nothing in the arithmetic justifies that. The pattern was already running before $2^0$, and it keeps running through it.

57. Take one expression, $\dfrac{a^3}{a^3}$, and read it two legitimate ways. The quotient law gives $a^{3-3} = a^0$. Canceling all three pairs gives $\dfrac{a \cdot a \cdot a}{a \cdot a \cdot a} = 1$, using nothing but the fact that a factor over itself is $1$. Since one expression has one value, $a^0$ and $1$ are the same number. No convention was adopted; the value was the only one available. The restriction $a \neq 0$ enters because both readings are readings of a quotient with $a^3$ in the denominator — at $a = 0$ that denominator is $0$ and the expression names nothing, so the argument says nothing about $0^0$.

58. Two mistakes. First, the student read the negative exponent as producing a negative number; a negative exponent produces a reciprocal, and all the values in the pattern below $2^0$ are positive. Second, the student multiplied the base by the exponent, treating $5^{-2}$ as $5 \cdot (-5)$ or similar, rather than counting factors. Correct: $5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25} = 0.04$. The expression that really equals $-25$ is $-5^2$.

59. The exponent counts factors; it is not the answer. An exponent of $0$ says no factors are being multiplied, and the value that leaves is $1$, not $0$. Correct: $x^0 = 1$ for every $x \neq 0$. The quotient that forces it is $\dfrac{x^3}{x^3}$, which is $x^{3-3} = x^0$ by the quotient law and $1$ by canceling — two readings of one expression, so $x^0 = 1$.

60. Two patterns disagree at $0^0$. Along the exponents, $a^0 = 1$ for every nonzero base, which argues for $1$. Along the bases, $0^n = 0$ for every positive exponent — $0^3 = 0$, $0^2 = 0$, $0^1 = 0$ — which argues for $0$. No value satisfies both. And the derivation of $a^0 = 1$ cannot break the tie, because it read the quotient $\dfrac{a^3}{a^3}$ two ways, and at $a = 0$ the denominator $0^3$ is $0$, so that quotient does not exist. With no pattern forcing an answer, $0^0$ is left undefined rather than assigned one by preference.

Exit ticket 10.3

61. $12^0 = 1$. $3^{-4} = \dfrac{1}{3^4} = \dfrac{1}{81}$. $(-2)^{-4} = \dfrac{1}{(-2)^4} = \dfrac{1}{16}$ — positive, because the base is raised to an even power.
62. $\dfrac{5x^{-3}}{y^{-2}} = \dfrac{5y^2}{x^3}$. The $x^{-3}$ moves down and the $y^{-2}$ moves up; the coefficient $5$ does not move.
63. $\dfrac{a^2b^{-5}}{a^{-3}b^{-1}} = a^{2-(-3)}b^{-5-(-1)} = a^5b^{-4} = \dfrac{a^5}{b^4}$
64. $2^0 = \mathbf{1}$ and $2^{-1} = \mathbf{\tfrac12}$. The single rule is: step down one exponent, divide by $2$. It produces $4$ from $8$, $2$ from $4$, $1$ from $2$, and $\tfrac12$ from $1$, without changing at any row.
65. Read $\dfrac{a^3}{a^3}$ two ways: the quotient law gives $a^{3-3} = a^0$, and canceling all three matched pairs gives $1$. One expression cannot have two values, so $a^0$ must be $1$ — the value is forced by a law already derived, not adopted by agreement.

Lesson 10.4 — Powers of Bases

Guided practice

66. $(a^2)^3 = (a \cdot a)(a \cdot a)(a \cdot a)$. The brace counts 3 groups of 2 factors, so 6 factors. The single power is $a^6$.
67. $(x^4)^3$ is 3 groups of 4 factors each, so $4 \cdot 3 = 12$ factors and the single power is $x^{12}$.
68. $(2^3)^2 = 8^2 = 64$, and $2^6 = 64$. The agreement confirms that multiplying the exponents produced the right number — the count of groups times factors per group is genuinely the total factor count, not just a plausible-looking rule.
69. $(2x)^3 = (2x)(2x)(2x)$, then regrouping gives $(2 \cdot 2 \cdot 2)(x \cdot x \cdot x)$, which is $8x^3$. The boxed law is $(ab)^n = a^nb^n$. The regrouping step is the derivation, and it is allowed because multiplication can be reordered.
70. $\left(\dfrac{x}{y}\right)^4 = \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y}$, then multiplying across gives $\dfrac{x \cdot x \cdot x \cdot x}{y \cdot y \cdot y \cdot y} = \dfrac{x^4}{y^4}$. The boxed law is $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$, for $b \neq 0$. The restriction is there because $b$ sits in a denominator throughout, and a denominator of zero names no number.
71. $a^2 \cdot a^3 = a^5$: two separate rows of factors are laid end to end, so the two counts add. $(a^2)^3 = a^6$: one row of two factors is repeated three times, so the answer is $3$ groups $\times$ $2$ factors per group, and the counts multiply. Different pictures give different arithmetic, and writing the factors out tells you which picture you have every time.

Independent practice

72. a) $(m^5)^4 = m^{5 \cdot 4} = m^{20}$ b) $(2^4)^3 = 2^{4 \cdot 3} = 2^{12}$ (which is $4096$) c) $(y^7)^2 = y^{14}$ d) $\big((a^2)^3\big)^2 = (a^6)^2 = a^{12}$ — or all at once, $a^{2 \cdot 3 \cdot 2} = a^{12}$
73. a) $(3x)^4 = 3^4x^4 = 81x^4$ — the coefficient takes the exponent too b) $(-2a)^5 = (-2)^5a^5 = -32a^5$; negative because the exponent is odd c) $(5m^2n)^3 = 5^3(m^2)^3n^3 = 125m^6n^3$ d) $(-4p^3q^2)^2 = (-4)^2(p^3)^2(q^2)^2 = 16p^6q^4$; positive because the exponent is even
74. a) $\left(\dfrac{x}{3}\right)^3 = \dfrac{x^3}{3^3} = \dfrac{x^3}{27}$ b) $\left(\dfrac{2a}{b}\right)^4 = \dfrac{(2a)^4}{b^4} = \dfrac{16a^4}{b^4}$ c) $\left(\dfrac{m^3}{n^2}\right)^5 = \dfrac{m^{15}}{n^{10}}$ d) $\left(\dfrac{3x^2}{y^4}\right)^2 = \dfrac{3^2x^4}{y^8} = \dfrac{9x^4}{y^8}$
75. a) $(x^4)^{-3} = x^{-12} = \dfrac{1}{x^{12}}$ b) $(2m)^{-3} = 2^{-3}m^{-3} = \dfrac{1}{8m^3}$ — the coefficient took the negative exponent as well, so $2^{-3} = \tfrac18$ ended up underneath c) $\left(\dfrac{a}{b}\right)^{-2} = \dfrac{a^{-2}}{b^{-2}} = \dfrac{b^2}{a^2}$ — a negative exponent on a fraction flips it d) $(3x^{-2})^{-1} = 3^{-1}(x^{-2})^{-1} = \dfrac{1}{3} \cdot x^2 = \dfrac{x^2}{3}$
76. Volume $= (2x^3)^3 = 2^3(x^3)^3 = 8x^9$ cubic centimeters. Two laws: power of a product, to send the outer exponent to both the $2$ and the $x^3$, and power of a power, to turn $(x^3)^3$ into $x^9$.
77. A square kilometer is $(10^3)^2 = 10^{3 \cdot 2} = 10^6$ square meters, which is $1{,}000{,}000$. The law is power of a power. (This is why doubling a length quadruples an area — the exponent on the length gets multiplied, not the length itself.)
78. $(ab)^n = a^nb^n$ is true because the derivation reorders factors: $(ab)(ab)(ab)$ can be regrouped as $(a \cdot a \cdot a)(b \cdot b \cdot b)$ since multiplication may be done in any order. In $a + b$ there are no factors to reorder, so that step has nothing to act on and the analogous claim fails. Counterexample 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 retires the claim permanently. (What $(a+b)^2$ actually equals is Chapter 12's work.)
79. The student sent the exponent to the $x$ but not to the $3$. Every factor inside the parentheses takes the exponent, coefficient included: $(3x)^2 = 3^2x^2 = 9x^2$. At $x = 1$: the correct answer gives $9$, and the student's $3x^2$ gives $3$.
80. The student used the product law, adding $3 + 4 = 7$, on an expression that is a power of a power. Correct: $(x^3)^4 = x^{3 \cdot 4} = x^{12}$. The count is $4$ groups of $3$ factors, so twelve factors in all — not seven.
81. a) $\left(x^4\right)^5 = x^{20}$, since $4 \cdot 5 = 20$, so the missing exponent is $\mathbf{4}$ b) $\left(2a^3\right)^3 = 2^3a^9 = 8a^9$, so the missing exponent is $\mathbf{3}$; both the coefficient and the variable confirm it, since $2^3 = 8$ and $3 \cdot 3 = 9$ c) $\left(\dfrac{m^4}{n}\right)^3 = \dfrac{m^{12}}{n^3}$, so the missing exponents are $\mathbf{12}$ and $\mathbf{3}$

Exit ticket 10.4

82. $(a^6)^3 = a^{18}$
83. $(-3m^2n^5)^3 = (-3)^3(m^2)^3(n^5)^3 = -27m^6n^{15}$; negative because the exponent is odd
84. $\left(\dfrac{2x^3}{y^2}\right)^4 = \dfrac{2^4x^{12}}{y^8} = \dfrac{16x^{12}}{y^8}$
85. $x^3 \cdot x^5 = x^8$: two rows of factors are set end to end, so the counts add — three copies of $x$ followed by five copies is eight copies. $(x^3)^5 = x^{15}$: one row of three factors is repeated five times, so the counts multiply — five groups of three is fifteen factors. In the first, factors are being joined; in the second, a group is being repeated.

Lesson 10.5 — Multivariable Expressions and Ratios of Monomials

Guided practice

86. One step to the right multiplies the value by $10$; one step to the left divides it by $10$ — the same single pattern in both directions, which is what makes the negative side of the scale no different in kind from the positive side. At $10^0$ the value is $1$, marked on the figure, and the everyday length labeled there is one meter, one long stride.
87. $\dfrac{10^3}{10^{-3}} = 10^{3-(-3)} = 10^6$. A kilometer is one million times as long as a millimeter — six orders of magnitude, which is the six ticks between the two callouts.
88. $\dfrac{15x^5y^3}{5x^2y^7} = 3x^{5-2}y^{3-7} = 3x^3y^{-4} = \dfrac{3x^3}{y^4}$. The base $y$ had the larger exponent underneath, so its leftover factors really are in the denominator.
89. $\dfrac{(2a^3)^4}{8a^5} = \dfrac{2^4a^{12}}{8a^5} = \dfrac{16a^{12}}{8a^5} = 2a^{12-5} = 2a^7$. Clearing the outer exponent must come first, and it has to reach the $2$ as well as the $a^3$: $2^4 = 16$, not $2$.
90. $\dfrac{6m^4n^{-2}}{9m^{-1}n^3} = \dfrac{6}{9}m^{4-(-1)}n^{-2-3} = \dfrac{2}{3}m^5n^{-5} = \dfrac{2m^5}{3n^5}$. On the base $m$ the subtraction is $4 - (-1) = 4 + 1 = 5$; writing the parentheses is what keeps it from being read as $4 - 1 = 3$.
91. A monomial is a number, a variable, or a product of numbers and variables — for example $7$, $x$, $-3a^2b$. $\dfrac{3x^2}{y}$ is not a monomial, because a variable appears in a denominator, which makes it a quotient of two monomials rather than a single product. It is, however, exactly the kind of thing that a simplified ratio of monomials comes out as.

Independent practice

92. a) $\dfrac{x^8y^2}{x^3y^6} = x^{8-3}y^{2-6} = x^5y^{-4} = \dfrac{x^5}{y^4}$ b) $\dfrac{24a^5b^3}{18a^2b^7} = \dfrac{24}{18}a^{5-2}b^{3-7} = \dfrac{4}{3}a^3b^{-4} = \dfrac{4a^3}{3b^4}$ c) $\dfrac{-14m^3n^8}{7m^7n^2} = -2m^{3-7}n^{8-2} = -2m^{-4}n^6 = -\dfrac{2n^6}{m^4}$ d) $\dfrac{p^4q^{-3}r}{p^{-2}q^2r^5} = p^{4-(-2)}q^{-3-2}r^{1-5} = p^6q^{-5}r^{-4} = \dfrac{p^6}{q^5r^4}$

93. a) $(3x^2y)^3 \cdot 2xy^4 = 27x^6y^3 \cdot 2xy^4 = 54x^{6+1}y^{3+4} = 54x^7y^7$. Clear the outer exponent first — $3^3 = 27$ — then use the product law. b) $\dfrac{(2m^3)^4}{(4m^2)^2} = \dfrac{16m^{12}}{16m^4} = m^{12-4} = m^8$. The coefficients happen to cancel exactly, which is easy to miss if the outer exponents are not both cleared first. c) $\left(\dfrac{a^4b}{a^2b^3}\right)^3 = (a^2b^{-2})^3 = a^6b^{-6} = \dfrac{a^6}{b^6}$. Simplifying inside first is much shorter than cubing top and bottom and dividing afterward, though both give the same answer. d) $\dfrac{5x^{-2}y^3}{20x^4y^{-1}} = \dfrac{5}{20}x^{-2-4}y^{3-(-1)} = \dfrac{1}{4}x^{-6}y^4 = \dfrac{y^4}{4x^6}$

94. a) $\dfrac{18r^6s^{-2}t^4}{12r^2s^3t^4} = \dfrac{3}{2}r^{6-2}s^{-2-3}t^{4-4} = \dfrac{3}{2}r^4s^{-5}t^0 = \dfrac{3r^4}{2s^5}$. The base $t$ vanished because $t^0 = 1$. b) $\dfrac{(2rs^2)^3t}{4r^2s^4t^{-2}} = \dfrac{8r^3s^6t}{4r^2s^4t^{-2}} = 2r^{3-2}s^{6-4}t^{1-(-2)} = 2rs^2t^3$ c) $\dfrac{-x^3y^5z^2}{x^5y^2z^{-1}} = -x^{3-5}y^{5-2}z^{2-(-1)} = -x^{-2}y^3z^3 = -\dfrac{y^3z^3}{x^2}$ d) $\dfrac{(3a^2b)^2}{(3ab^2)^2} = \dfrac{9a^4b^2}{9a^2b^4} = a^{4-2}b^{2-4} = a^2b^{-2} = \dfrac{a^2}{b^2}$

95. Front numbers: $\dfrac{3.4}{1.7} = 2$. Exponents: $\dfrac{10^8}{10^4} = 10^{8-4} = 10^4$. So the ratio is $2 \times 10^4 = 20{,}000$: about twenty thousand times as many people live in the country as in the town.

96. $\dfrac{10^6}{10^{-3}} = 10^{6-(-3)} = 10^9$. The Richmond-to-Chicago distance is about a billion times the thickness of a credit card, and the two lengths are nine orders of magnitude apart — the nine ticks between $10^{-3}$ and $10^6$ on the scale.

97. $\dfrac{4 \times 10^9}{2 \times 10^6} = 2 \times 10^{9-6} = 2 \times 10^3$, which is $2000$ photos.

98. $\dfrac{(2x^3)^4}{8x^5}$. Step 1, power of a product: the outer exponent reaches both factors, giving $2^4(x^3)^4$. Step 2, power of a power: $(x^3)^4 = x^{12}$, so the numerator is $16x^{12}$. Step 3, quotient law: divide the coefficients, $16 \div 8 = 2$, and subtract the exponents, $12 - 5 = 7$. Result: $2x^7$.

99. $\dfrac{x^3}{x^7} = x^{3-7} = x^{-4}$ by the quotient law, and $x^{-4} = \dfrac{1}{x^4}$ by the negative-exponent law — which itself came from counting the four unmatched factors left below the bar in $\dfrac{x^3}{x^7}$. So all three are the same quantity at every $x \neq 0$. This chapter reports $\dfrac{1}{x^4}$, because the convention is that a final answer contains no negative exponent; the form with a positive exponent also shows at a glance that the value is a small positive number when $x$ is large.

100. On the base $b$ the student subtracted in the wrong direction, computing $7 - 3 = 4$ instead of $3 - 7 = -4$, and then left the result in the numerator. The subtraction is always numerator exponent minus denominator exponent, and a negative result means the leftover factors are in the denominator. Correct: $\dfrac{24a^5b^3}{18a^2b^7} = \dfrac{4}{3}a^3b^{-4} = \dfrac{4a^3}{3b^4}$. A check at $a = 1$, $b = 1$ would not catch this; at $a = 1$, $b = 2$ the original is $\tfrac{24 \cdot 8}{18 \cdot 128} = \tfrac{1}{12}$, the correct answer gives $\tfrac{4}{3 \cdot 16} = \tfrac{1}{12}$, and the student's gives $\tfrac{4 \cdot 16}{3} = \tfrac{64}{3}$.

101. The student subtracted the exponents as $4 - 1$ instead of $4 - (-1)$, dropping the sign on the denominator's exponent. Subtracting a negative adds: $4 - (-1) = 4 + 1 = 5$. Correct: $\dfrac{6m^4}{9m^{-1}} = \dfrac{2}{3}m^5 = \dfrac{2m^5}{3}$. Written another way, $\dfrac{1}{m^{-1}} = m$, so the denominator's $m^{-1}$ contributes an extra factor of $m$ upstairs.

102. At $x = 2$, $y = 2$: the original is $\dfrac{2^8 \cdot 2^2}{2^3 \cdot 2^6} = \dfrac{256 \cdot 4}{8 \cdot 64} = \dfrac{1024}{512} = 2$, and the simplified form is $\dfrac{2^5}{2^4} = \dfrac{32}{16} = 2$. At $x = 3$, $y = 1$: the original is $\dfrac{3^8 \cdot 1}{3^3 \cdot 1} = \dfrac{6561}{27} = 243$, and the simplified form is $\dfrac{3^5}{1} = 243$. All four values: $2$, $2$, $243$, $243$.

     Agreement at two pairs is not a proof because two expressions can agree at finitely many points and differ elsewhere — the check confirms nothing about the infinitely many pairs not tried. But a disagreement at even one allowed pair would be a proof of error, because a correct simplification has to equal the original at every allowed value; a single counterexample is enough to rule it out. That asymmetry is why the check is worth thirty seconds even though it proves nothing on its own. (Testing $x = 1$ alone would be a poor check, since $1$ to any power is $1$ and most exponent errors survive it.)