module 2 – exponentsembaweb/qsw/module_2_exponents.pdfmodule 2 – exponents 3 exponents...

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Quantitative Skills WorkshopModule 2 – Exponents

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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

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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.



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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

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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



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Exponents: BasicsExponents: BasicsExercisesExercises

1. y(x–1) =

2. (x1/2)(y1/2)

3. (1/9)–2 =

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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 =



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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.

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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…



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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.

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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.



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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

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Summary of Exponent RulesSummary of Exponent Rules

baba xxx += bab

a

xxx −=

abba xx =)( b aba xx =/

x–a = 1/xax0 = 1x1 = x



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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

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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.



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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:

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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




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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


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DecayDecayExerciseExercise

The general formula we derived from the decay application is:The general formula we derived from the decay application is:


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?



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DecayDecayExercise: **Answer**Exercise: **Answer**


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.

module 2 – exponentsembaweb/qsw/module_2_exponents.pdfmodule 2 – exponents 3 exponents...

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