Page 2© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
3
ExponentsExponentsDefinitionsDefinitions
• If a variable, x, is raised to the power of some number, say 2, then we call “x” the base and “2” the exponent.
• The exponent term tells you how many times the base is multiplied by itself.
x2 = x * x
x3 = x * x * x
24 = 2*2*2*2 = 16
4
Exponents in UseExponents in Use
• There are many applications of exponents, particularly in finance. For example, in calculating compound interest, the number of years of compounding and the number of times per year the interest is compounded are both exponents in the formula for the value of the investment over time.
• In this module we will cover the basic rules or laws of exponents. During the in-residence QSW lectures, we will cover compound interest and other advanced applications of exponents.
Page 3© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
5
Negative ExponentsNegative Exponents
• For a negative exponent, take one over the base (the reciprocal of the base) and raise it to the absolute value of the exponent:
x–20 = 1/x20
2–3 = 1/23 = 1/(2*2*2) = 1/8 = 0.125
100–2 = 1/1002 = 1/10,000 = 0.0001
(4/5)–2 = 1/(4/5)2 = (1/1) * (5/4)2 = (5/4)2 = 1.5625
6
Fractional ExponentsFractional Exponents
• In the case of an exponent that is a fraction, you need to take a root. For example, take a square root if the exponent is 1/2 or a cube root if the exponent is 1/3.
Examples:
32727
24433/1
2/1
2/1
==
==
= xx
Page 4© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
7
Exponents: BasicsExponents: BasicsExercisesExercises
1. y(x–1) =
2. (x1/2)(y1/2)
3. (1/9)–2 =
8
Exponents: BasicsExponents: BasicsExercises: **Answers**Exercises: **Answers**
1. y(x–1) = y/x
2.
3. (1/9)–2 = 1/(1/9)2 = 92 = 81
))(())(( 2/12/1 yxyx =
Page 5© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
9
Rule of “1”Rule of “1”
• A number raised to the power of 1 is equal to itself:
x1 = x
21 = 2
1001 = 100
• The logic behind this rule is that the power “1” tells you to multiply the base (2 or 100 or x) by itself only one time. That means that you get the same number back.
10
Rule of “0”Rule of “0”
• A number raised to the power of 0 is equal to one:
x0 = 1
20 = 1
1000 = 1
• This is just one of those rules that you need to know…
Page 6© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
11
Product RuleProduct Rule
• To multiply numbers that have the same base but different exponents, you add the exponents:
(x1)(x4) = x(1+4) = x5
(22) * (23) = 2(2+3) = 25 = 2*2*2*2*2 = 32
(1000)(1002) = 100(0+2) = 1002 = 10,000 (or, knowing that 1000 = 1, you could calculate this directly as 1002)
• As you can see, this is just a convenient shortcut. You could do the calculation differently (e.g., multiply 22 = 4 by 23 = 8 to get 32). However, when working with long series of numbers or when working with variables rather than numbers, it is a convenient shortcut to know.
12
Power RulePower Rule
• If there is one base term and two exponents, you multiply the exponents:
(x0.06)100 = x0.06*100 = x6
(22)3 = 22*3 = 26 = 64
(103.5)2 = 103.5*2 = 107 = 10 million
Note that it is not true that (x + y)2 = x2 + y2. If you run across an expression like this you have to multiply it out. We’ll do this in the next set of exercises.
Page 7© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
13
Quotient RuleQuotient Rule
• To divide numbers that have the same base but two exponents, you subtract the exponents:
xa ÷ xb = x(a-b)
26 ÷ 22 = 2(6-2) = 24 = 16
801.5 / 800.5 = 80(1.5-0.5) = 801 = 80
14
Summary of Exponent RulesSummary of Exponent Rules
baba xxx += bab
a
xxx −=
abba xx =)( b aba xx =/
x–a = 1/xax0 = 1x1 = x
Page 8© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
15
Exponents RulesExponents RulesExercisesExercises
1. 1560 =
2. (106)0.5
3. (32)*(33) =
4. 2251.5 / 225 =
5. Show that (x + 5)2 does not equal x2 + 52
16
Exponent RulesExponent RulesExercises: **Answers**Exercises: **Answers**
1. 1560 = 1
2. (106)0.5 = 10(6*0.5) = 103 = 1,000
3. (32)*(33) = 3(2+3) = 35 = 243
4. 2251.5 / 225 = 225( 1.5 – 1) = 2250.5 = 15
5. Show that (x + 5)2 ≠ x2 + 52
(x + 5)2 = (x + 5)(x + 5) = x(x + 5) + 5(x + 5)= (x2 + x•5) + (5•x + 52)= x2 + 10x + 25. This is not the same as x2 + 25.
The difference is the middle term, 10x.
Page 9© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
17
Example: If a chemical company has N=1000 customers and Example: If a chemical company has N=1000 customers and retains half of those each year, how many customers does it retains half of those each year, how many customers does it
have left after 1year? 2 years? 3 years? T years?have left after 1year? 2 years? 3 years? T years?
NTT= ×1000 0 50.After After TT periods:periods:
12550.0100050.050.050.01000 33 =×=×××=NAfter After 33 periods:periods:
N221000 0 50 0 50 1000 0 50 250= × × = × =. . .After After 22 periods:periods:
Application: Decay of a Customer BaseApplication: Decay of a Customer Base
N1 1000 0 50 500= × =.After After 11 period:period:
18
NTT= ×1000 0 50.After After TT periods:periods:
Application: Decay of a Customer BaseApplication: Decay of a Customer Base
The general formula isThe general formula is::
Amount remaining = (base amt)*(fraction retained each year)Amount remaining = (base amt)*(fraction retained each year)# years# years
Example: If a chemical company has N=1000 customers and Example: If a chemical company has N=1000 customers and retains half of those each year, how many customers does it retains half of those each year, how many customers does it
have left after 1year? 2 years? 3 years? T years?have left after 1year? 2 years? 3 years? T years?
Page 10© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
19
0
200
400
600
800
1000
Number
0 1 2 3 4 5 6 7 8 9 10 Period
What This Looks Like
Application: Decay of a Customer BaseApplication: Decay of a Customer BaseExample: If a chemical company has N=1000 customers and Example: If a chemical company has N=1000 customers and retains half of those each year, how many customers does it retains half of those each year, how many customers does it
have left after 1year? 2 years? 3 years? T years?have left after 1year? 2 years? 3 years? T years?
20
DecayDecayExerciseExercise
The general formula we derived from the decay application is:The general formula we derived from the decay application is:
Amount remaining = (base amt)*(fraction retained each year)Amount remaining = (base amt)*(fraction retained each year)# years# years
1.1. The price of a new car is $30,000 and it depreciates at a rate The price of a new car is $30,000 and it depreciates at a rate of 15% per year. (This is, of course, a simplification: the of 15% per year. (This is, of course, a simplification: the depreciation rate for cars is not a constant.) What is the depreciation rate for cars is not a constant.) What is the value of the car at the end of the first year? value of the car at the end of the first year?
2.2. What is the value of the car at the end of the second year? What is the value of the car at the end of the second year? At the end of the fifth year?At the end of the fifth year?
3.3. What is the general decay formula in this case?What is the general decay formula in this case?
Page 11© 2002 The Regents of The University of MichiganAll Rights Reserved. No duplication without prior, written consent.
Quantitative Skills WorkshopModule 2 – Exponents
21
DecayDecayExercise: **Answer**Exercise: **Answer**
Amount remaining = (base amt)*(fraction retained each year)Amount remaining = (base amt)*(fraction retained each year)# years# years
First calculate the fraction retained each year: since 15% is loFirst calculate the fraction retained each year: since 15% is lost st each year, the fraction retained is (1 each year, the fraction retained is (1 –– 0.15) = 0.85 or 85%.0.15) = 0.85 or 85%.
1.1. Value after 1 year = 30,000(0.85)Value after 1 year = 30,000(0.85)11 = 30,000(0.85) = $25,500= 30,000(0.85) = $25,500
2.2. Value after 2 years = 30,000(0.85)Value after 2 years = 30,000(0.85)22 = 30,000(0.7225) = $21,675= 30,000(0.7225) = $21,675
Value after 5 years = 30,000(0.85)Value after 5 years = 30,000(0.85)55 = 30,000(0.4437) = $13,311= 30,000(0.4437) = $13,311
3.3. The general formula is: 30,000(0.85)The general formula is: 30,000(0.85)TT, where T is the number , where T is the number of years since the purchase of the car.of years since the purchase of the car.