21
NEGATIVE EXPONENTS
Power of a fraction
"To raise a fraction to a power, raise the numerator
and denominator to that power."
| Example 1. |  |
For, according to the meaning of the exponent, and the rule for multiplying fractions:
| Example 2. Apply the rules of exponents: |  |
Solution. We must take the 4th power of everything. But to take a power of a power -- multiply the exponents:
Problem 1. Apply the rules of exponents.
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Do the problem yourself first!
| a) |  |
= | x² y² |
b) |  |
= | 8x³ 27 |
c) |  |
= | 
|
| d) |  |
= | 
|
| e) |  |
= | x² − 2x + 1 x² + 2x + 1 |
Perfect Square Trinomial |
Subtracting exponents
Shortly (Lesson 20), we will see the following rule for canceling:
| ax ay |
= | x y |
"If the numerator and denominator have a common factor,
it may be canceled."
Consider these examples of canceling:
| 2· 2· 2· 2· 2 2· 2 |
= | 2· 2· 2 |
| ___2· 2___ 2· 2· 2· 2· 2 |
= | __1__ 2· 2· 2 |
If we write these examples with exponents, then
 22 |
= | 23 |
| 2² 25 |
= | 1 23 |
In each case, we subtract the exponents. But when the exponent in the denominator is larger, we write 1-over their difference.
| Example 3. |  x3 |
= | x5 |
 x8 |
= | 1 x5 |
|
Here is the rule:
Problem 2. Simplify the following. (Do not write a negative exponent.)
| a) |  |
= | x3 | b) | x² x5 |
= | 1 x3 |
c) | x x5 |
= | 1 x4 |
| d) | x² x |
= | x | e) |  |
= | −x4 | f) |  |
= | 1 x² |
Problem 3. Simplify each of the following. Then calculate each number.
| a) |  |
= | 23 | = | 8 | b) | 2² 25 |
= | 1 23 |
= | 1 8 |
c) | 2 25 |
= | 1 24 |
= | 1 16 |
| d) | 2² 2 |
= | 2 | e) |  |
= | −24 | = | −16 | f) |  |
= | 1 2² |
= | 1 4 |
| Example 4. Simplify by reducing to lowest terms: |  |
Solution. Consider each element in turn:
Problem 4. Simplify by reducing to lowest terms. (Do not write negative exponents.
| a) |  |
= | y³ 5x³ |
b) |  |
= | − | 8a³ 5b³ |
| c) |  |
= | − | 3z_ 5x4y3 |
d) |  |
= | c³ 16 |
| e) | (x + 1)³ (x − 1) (x − 1)³ (x + 1) |
= | (x + 1)² (x − 1)² |
Negative exponents
We are now going to extend the meaning of an exponent to more than just a positive whole number. We will do that in such a way that the usual rules of exponents will hold. That is, we will want the following rules to hold for any numbers: positive, negative, 0 -- even fractions!
| aman | = | am + n | Same Base |
| (ab)n | = | anbn | Power of a Product |
| (am)n | = | amn | Power of a Power |
We begin by defining a number raised to a negative exponent to be the reciprocal of that power with a positive exponent.
| a−n | = | 1 an |
a−n is the reciprocal of an.
| Example 5. | 2−3 | = | 1 23 |
= | 1 8 |
The base, 2, does not change. The negative exponent becomes positive -- in the denominator.
Example 6. Compare the following three numbers. That is, evaluate them:
3−2 −3−2 (−3)−2
| Answers. | 3−2 | = | 1 32 |
= | 1 9 |
−3−2 is the negative of 3−2. The base is still 3.
| −3−2 | = − | 1 9 |
As for (−3)−2, the base here is −3:
| (−3)−2 | = | 1 (−3)2 |
= | 1 9 |
| Example 7. Simplify | a² a5 |
. |
Solution. Since we have invented negative exponents, we can now subtract any exponents as follows:
| a² a5 |
= a2 − 5 = a−3 |
That is, we now have the following rule for any numbers m, n:
| In fact, we defined a− n as | 1 an |
, because we want that rule |
to hold. We want
 |
= | a−3 |
But
|
= | 1 a3 |
| Therefore, we define a−3 as | 1 a3 |
. |
| Example 8. | a−1 | = | 1 a |
a−1 is now a symbol for the reciprocal, or multiplicative inverse, of any number a. It appears in the following rule (Lesson 6):
a· a−1 = 1
Problem 5.
| a) | (log 2)(log 2)−1 = 1 | b) | (x² − 7x + 5)· (x² − 7x + 5)−1 = 1 |
| c) ( | 2 3 |
)−1 | = | 3 2 |
Example 9. Use the rules of exponents to evaluate (2−3· 104)−2.
| Solution. | (2−3· 104)−2 | = | 26· 10−8 | Power of a power |
| = |  |
|||
| = | _64_ 100,000,000 |
|||
Problem 6. Evaluate the following.
| a) | 2−4 | = |
1 24 |
= |
1 16 |
b) | 5−2 | = |
1 52 |
= |
1 25 |
c) | 10−1 | = |
1 101 |
= |
1 10 |
| d) | (−2)−3 | = |
1 (−2)3 |
= |
1 −8 |
= | − |
1 8 |
| e) | (−2)−4 | = |
1 (−2)4 |
= |
1 16 |
f) | −2−4 | = | − |
1 24 |
= | − |
1 16 |
g) (½)−1 = 2. 2 is the reciprocal of ½.
Problem 7. Use the rules of exponents to evaluate the following.
| a) | 10²· 10−4 = 102 − 4 = 10−2 = 1/100. |
| b) | (2−3)² | = | 2−6 | = | 1 26 |
= |
1 64 |
| c) | (3−2· 24)−2 | = | 34· 2−8 | = |
34 28 |
= |
81 256 |
| d) | 2−2· 2 | = | 2−2+1 | = | 2−1 | = |
1 2 |
Problem 8. Rewrite without a denominator.
| a) | x² x5 |
= | x2−5 | = | x−3 | b) | y y6 |
= | y1−6 | = | y−5 |
| c) |  |
= | x−3y−4 | d) |  |
= | a−1b−6c−7 |
| e) | 1 x |
= | x−1 | f) | 1 x3 |
= | x−3 |
| g) | (x + 1) x |
= | (x + 1)x−1 | h) |  |
= | (x + 2)−4 |
| * | * | * |
Reciprocals come in pairs. a−n is the reciprocal of an. And an is the reciprocal of a−n:
| 1 a−n |
= | an |
Together, these imply:
Factors may be shifted between the numerator and denominator
by changing the sign of the exponent.
| Example 10. Rewrite without a denominator: |  |
| Answer. | |
= | 10−3 + 5 − 2 + 4 | = | 104 | = | 10,000 |
Exponent 2 goes into the numerator as −2; exponent −4 goes there as +4.
Problem 9. Rewrite without a denominator and evaluate.
| a) | 2² 2−3 |
= 22 + 3 = 25 = 32 | b) | 10² 10−2 |
= 102 + 2 = 104 = 10,000 |
| c) |  |
= 102 − 5 − 4 + 6 = 10−1 = | 1 10 |
| d) |  |
= 25 − 6 + 9 − 7 = 21 = 2 |
Problem 10. Rewrite with positive exponents only.
| a) | x y−2 |
= | xy² | b) |  |
= |  |
c) |  |
= |  |
| d) |  |
= |  |
e) |  |
= |  |
Problem 11. Apply the rules of exponents, then rewrite with positve exponents.
| a) |  |
= |  |
= |  |
b) |  |
= |  |
= |  |
Next Lesson: Multiplying and dividing algebraic fractions
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