13
EXPONENTS
WHEN A NUMBER is repeatedly multiplied by itself, we get the powers of that number (Lesson 1).
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.
Problem 2. What does each symbol mean?
| a) | x5 = xxxxx | b) | 53 = 5· 5· 5 | |
| c) | (5a)3 = 5a· 5a· 5a | d) | 5a3 = 5aaa | |
In part c), the parentheses indicate that 5a is the base. In part d), 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? 4. It indicates the power.
Problem 4. Write out the meaning of these symbols.
| a) | a²a3 = aa· aaa | b) | (ab)3 = ab· ab· ab | c) | (a²)3 = a²· a²· a² |
Problem 5. Write out the meaning of these symbols. In each one, what is the base?
a) a4 = aaaa. The base is a.
| b) | −a4 | = | −aaaa. The base again is a. This is the negative of a4. A minus sign always signifies the negative of the number that follows. −5 is the negative of 5. And −a4 is the negative of 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.
| a) | xxxxxx = x6 | b) | xxyyyy = x²y4 | c) | xyxxyx = x4y² | d) | xyxyxy = x3y3 |
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
aman = am + 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) | xmynxpyq = xm + pyn + q | l) | apbqab = ap + 1bq + 1 | ||||
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 an issue.
Problem 11. 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 12. Apply the rule Same Base.
| a) | xnx² = xn + 2 | b) | xnx = xn + 1 | c) | xnxn = x2n | d) | xnx1 − n = x | |||
| e) | x· xn + 2 = xn + 3 | f) | xnxm = xn + m | g) | x2nx2 − n = xn + 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 13. Apply the rules of exponents.
| a) | (xy)4 = x4y4 | b) | (pqr)5 = p5q5r5 | c) | (2abc)3 = 23a3b3c3 |
| d) x3y²z4(xyz)5 | = | x3y²z4· x5y5z5 Rule 2, |
| = | x8y7z9 Rule 1. | |
Rule 3: Power of a Power
(am)n = amn
"To take a power of a power, multiply the exponents."
For example, (a²)3 = a2 · 3 = a6.
Why do we do that? Again, because of what the symbols mean:
(a²)3 = a²a²a² = a3 · 2 = a6
Problem 14. Apply the rules of exponents.
| a) | (x²)5 = x10 | b) | (a4)8 = a32 | c) | (107)9 = 1063 |
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 15. 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 16. Apply the rules of exponents.
| a) | 2x5y4(2x3y6)5 = 2x5y4· 25x15y30 = 26x20y34 |
b) abc9(a²b3c4)8 = abc9· a16b24c32 = a17b25c41
Problem 17. 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 18. Square the following.
| a) | x5 = x10 | b) | 8a3b6 = 64a6b12 | c) | −6x7 = 36x14 | d) | xn = x2n |
Note: In part c): The square of a negative number is positive.
(−6)(−6) = +36.
Problem 19. Apply a rule of exponents -- if possible.
| a) | x²x5 = x7, Rule 1. | b) | (x²)5 = x10, Rule 3. |
| c) x² + x5 | Not possible. The rules of exponents apply only to multiplication. |
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.
Problem 20. Apply the rules of exponents.
| a) | (xn)n = xn · n = xn² | b) | (xn)² = x2n |
Problem 21. Apply a rule of exponents or add like terms -- if possible.
a) 2x² + 3x4 Not possible. These are not like terms (Lesson 1).
b) 2x²· 3x4 = 6x6. Rule 1.
c) 2x3 + 3x3 = 5x3. Like terms. The exponent does not change.
d) x² + y² Not possible. These are not like terms.
e) x² + x² = 2x². Like terms.
f) x²· x² = x4. Rule 1.
g) x²· y3 Not possible. Different bases.
h) 2· 26 = 27. Rule 1.
i) 35 + 35 + 35 = 3· 35 (Like terms) = 36.
Next Lesson: Multiplying out. The distributive rule.
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