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A L G E B R A

EXPONENTS

WHEN All THE FACTORS OF A PRODUCT are equal -- 6· 6· 6· 6 -- we call the product a power of that factor (Lesson 1).

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Problem 1. What number is

a)	the third power of 2? 2· 2· 2 = 8	b)	the fourth power of 3? = 81

c)	the fifth power of 10? = 100,000	d)	the first power of 8? = 8

Now, rather than write the third power of 2 as 2· 2· 2, we write 2 just once -- and place an exponent: 2³. 2 is called the base. The exponent indicates the number of times to repeat the base as a factor.

The student must take care not to confuse 3a, which means 3 times a, with a³, which means a times a times a.

3a	=	a + a + a,
a³	=	a· a· a.

Problem 2. What does each symbol mean?

a)	x⁵ = xxxxx	b)	5³ = 5· 5· 5	c)	5· 3 = 3 + 3 + 3 + 3 + 3.

d)	(5a)³ = 5a· 5a· 5a	e)	5a³ = 5aaa

In part d), the parentheses indicate that the base is 5a. In part e), only a is the base. The exponent does not apply to 5.

Problem 3. 3⁴ = 81.

a) Which number is called the base? 3

b) Which number is the power? 81 is the power of 3.

c) Which number is the exponent?

Problem 4. Write out the meaning of these symbols.

a²a³ =

(ab)³ =

(a²)³ =

Problem 5. Write out the meaning of these symbols. In each one, what is the base?

a) a⁴ = aaaa. The base is a.

−a⁴

c) (−a)⁴ = (−a)(−a)(−a)(−a). Here, the base is (−a).

Problem 6. Evaluate.

a) 2⁴ = 16.

b) −2⁴ = −16. This is the negative of 2⁴. The base is 2. See
Problem 5b) above.

−a = (−1)a, for any number a. (Lesson 6.) Therefore −2⁴ = (−1)2⁴. And therefore according to the order of operations (Lesson 1), this is (−1)16 = −16.

c) (−2)⁴ = +16, according to the Rule of Signs (Lesson 4).
The parentheses indicate that the base is −2. See Problem 5c).

Example 1. Negative base.

(−2)³ = (−2)(−2)(−2) = −8,

again according to the Rule of Signs. Whereas,

(−2)⁴ = +16.

When the base is negative, and the exponent is odd, then the product is negative. But when the base is negative, and the exponent is even, then the product is positive.

Problem 7. Evaluate.

a)	(−1)² = 1	b)	(−1)³ = −1	c)	(−1)⁴ = 1	d)	(−1)⁵ = −1

e)	(−1)¹⁰⁰ = 1	f)	(−1)²⁵³ = −1	g)	(−2)⁴ = 16	h)	(−2)⁵ = −32

Problem 8. Rewrite using exponents.

xxxxxx

xxyyyy

xyxxyx

xyxyxy

Problem 9. Rewrite using exponents.

a)	(x + 1)(x + 1) = (x + 1)²	b)	(x − 1)(x − 1)(x − 1) = (x − 1)³

c)	(x + 1)(x − 1)(x + 1)(x − 1) = (x + 1)²(x − 1)²

d)	(x + y)(x + y)² = (x + y)³

Three rules

Rule 1. Same Base

a^maⁿ = a^{m + n}

"To multiply powers of the same base, add the exponents."

For example, a²a³ = a⁵.

Why do we add the exponents? Because of what the symbols mean. Problem 4a.

Example 2. Multiply 3x²· 4x⁵· 2x

Solution. The problem means (Lesson 5): Multiply the numbers, then combine the powers of x :

3x²· 4x⁵· 2x = 24x⁸

Two factors of x -- x² -- times five factors of x -- x⁵ -- times one factor of x, produce a total of 2 + 5 + 1 = 8 factors of x : x⁸.

Problem 10. Multiply. Apply the rule Same Base.

a)	5x²· 6x⁴ = 30x⁶	b)	7x³· 8x⁶ = 56x⁹	c)	x· 5x⁴ = 5x⁵

d)	2x· 3x· 4x = 24x³	e)	x³· 3x²· 5x = 15x⁶	f)	x⁵· 6x⁸y² = 6x¹³y²

g)	4x· y· 5x²· y³ = 20x³y⁴	h)	2xy· 9x³y⁵ = 18x⁴y⁶

i)	a²b³a³b⁴ = a⁵b⁷	j)	a²bc³b²ac = a³b³c⁴

k)	x^myⁿx^py^q = x^{m + p}y^{n+ q}	l)	a^pb^qab = a^{p + 1}b^{q + 1}

Problem 11. Distinguish the following:

x· x and x + x.

Example 3. Compare the following:

a) x· x⁵ b) 2· 2⁵

Solution.

a) x· x⁵ = x⁶

b) 2· 2⁵ = 2⁶

Part b) has the same form as part a). It is part a) with x = 2.

One factor of 2 multiplies five factors of 2 producing six factors of 2.

2· 2 = 4 is not correct here.

Problem 12. Apply the rule Same Base.

a)	xx⁷ = x⁸	b)	3· 3⁷ = 3⁸	c)	2· 2⁴· 2⁵ = 2¹⁰

d)	10· 10⁵ = 10⁶	e)	3x· 3⁶x⁶ = 3⁷x⁷

Problem 13. Apply the rule Same Base.

a)	xⁿx² = x^{n + 2}	b)	xⁿx = x^{n + 1}	c)	xⁿxⁿ = x²ⁿ	d)	xⁿx^{1 − n} = x

e)	x· 2x^{n − 1} = 2xⁿ	f)	xⁿx^m = x^{n + m}	g)	x²ⁿx^{2 − n} = x^{n + 2}

Rule 2: Power of a Product of Factors

(ab)ⁿ = aⁿbⁿ

"Raise each factor to that same power."

For example, (ab)³ = a³b³.

Why may we do that? Again, according to what the symbols mean:

(ab)³ = ab· ab· ab = aaabbb = a³b³.

The order of the factors does not matter:

ab· ab· ab = aaabbb.

Problem 14. Apply the rules of exponents.

(xy)⁴ =

(pqr)⁵ =

(2abc)³ =

d) x³y²z⁴(xyz)⁵	=	x³y²z⁴· x⁵y⁵z⁵ Rule 2,

	=	x⁸y⁷z⁹ Rule 1.

Rule 3: Power of a Power

(a^m)ⁿ = a^mn

"To take a power of a power, multiply the exponents."

For example, (a²)³ = a^2· 3 = a⁶.

Why do we do that? Again, because of what the symbols mean:

(a²)³ = a²a²a² = a^3· 2 = a⁶

Problem 15. Apply the rules of exponents.

(x²)⁵

(a⁴)⁸

(10⁷)⁹

Example 4. Apply the rules of exponents: (2x³y⁴)⁵

Solution. Within the parentheses there are three factors: 2, x³, and y⁴. According to Rule 2 we must take the fifth power of each one. But to take a power of a power, we multiply the exponents. Therefore,

(2x³y⁴)⁵ = 2⁵x¹⁵y²⁰

Problem 16. Apply the rules of exponents.

a)	(10a³)⁴ = 10,000a¹²	b)	(3x⁶)² = 9x¹²	c)	(2a²b³)⁵ = 32a¹⁰b¹⁵

d)	(xy³z⁵)² = x²y⁶z¹⁰	e)	(5x²y⁴)³ = 125x⁶y¹²

f) (2a⁴bc⁸)⁶ = 64a²⁴b⁶c⁴⁸

Problem 17. Apply the rules of exponents.

a)	2x⁵y⁴(2x³y⁶)⁵ = 2x⁵y⁴· 2⁵x¹⁵y³⁰ = 2⁶x²⁰y³⁴

b) abc⁹(a²b³c⁴)⁸ = abc⁹· a¹⁶b²⁴c³² = a¹⁷b²⁵c⁴¹

Problem 18. Use the rules of exponents to calculate the following.

a)	(2· 10)⁴ = 2⁴· 10⁴ = 16· 10,000 = 160,000

b) (4· 10²)³ = 4³· 10⁶ = 64,000,000

c) (9· 10⁴)² = 81· 10⁸ = 8,100,000,000

Example 5. Square x⁴.

Solution. (x⁴)² = x⁸.

Thus to square a power, double the exponent.

Problem 19. Square the following.

x⁵

8a³b⁶

−6x⁷

xⁿ

Part c) illstrates: The square of a number is always positive.

(−6)(−6) = +36. The Rule of Signs.

Except 0² = 0.

Problem 20. Apply a rule of exponents -- if possible.

x²x⁵

(x²)⁵

c) x² + x⁵

In summary: Add the exponents when the same base appears twice: x²x⁴ = x⁶. Multiply the exponents when the base appears once -- and in parentheses: (x²)⁵ = x¹⁰.