CHAPTER 4 Exponential and Logarithmic Functions Section 4.1 Exponential Functions An exponential function with base a has the form f ( x ) = a x where a is a constant, a > 0 and a 6= 1. The graph of an exponential function has a horizontal asymptote but does not have a vertical asymptote. Its characteristic shape is that of a a gradual bend. 8

x 3 2 1 0 1 2 3

6

4

2

3

2

1

2

1

y 1/8 1/4 1/2 1 2 4 8

3

Figure 4.1: Graph of f ( x ) = 2x on

3  x  3.

Using the standard tools of translation and reflection from the chapter on functions, the graph can be changed in a predictable fashion. Example 4.1 : Sketch the graph of f ( x ) = 2

x.

Solution: The change from 2x to 2 graph.

x

causes a reflection around the y axis, resulting in the following

Example 4.2 : Sketch the graph of f ( x ) = 4

2

x.

Solution: This function has three changes: • The change from 2x to 2

x

causes a reflection around the y axis.

• The negative sign in front of 2

x

causes a refelection over the x axis.

• Adding 4 to the function causes a vertical shift of 4 units. resulting in the following graph.

41

Bradshaw - Math 188

Chapter 4 Notes

42

8

6

4

2

3

2

1

2

1

Figure 4.2: Graph of f ( x ) = 2

x

on

3

3  x  3.

4

2

3

2

1

2

1

3

2

4 Figure 4.3: Graph of f ( x ) = 4

2

x

on

3x3

Solving Equations Solving exponential equations with “nice numbers” involves writing both sides of the equation with the same base. Example 4.3 : Solve 42x+5 = 64 Solution: For this problem, since 64 = 43 , we have 42x+5 = 64 42x+5 = 43 2x + 5 = 3 x=

1

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Chapter 4 Notes

43

Section 4.2 The Natural Exponential Function Financial Formulas A practical use of exponential functions can be found in the following three financial formulas. • Compound Interest Formula: 8 Used for single deposits. A = amount in the account > > > > > > ✓ ◆ < P = principal - the money you put in i kt i = annual interest rate A = P 1+ where > k > > k = periods per year > > > : t = number of years • Annuity Formula: Used for savings 8 accounts. A = amount in the account > 2✓ 3 ◆kt > > > i > > 17 6 1+ k < P = deposit each period 6 7 i = annual interest rate A = P6 7 where i > 4 5 > > k = periods per year > > k > : t = number of years • Loan Formula: Used for loans. 8 L > 2 ✓ ◆ kt 3 > > > i > P > 1+ 61 7 < k 6 7 i L = P6 7 where i > 4 5 > > k > > k > : t

= amount of the loan = payment each period = annual interest rate = periods per year = number of years

Example 4.4 : You have a car loan for 4 years, with monthly payments. The loan amount is $9600 and the annual interest rate is 8.1%. What are your monthly payments? Solution: Since you are making monthly payments, k = 12. This gives 2 ✓ ◆ kt 3 i 1+ 61 7 k 6 7 L = P6 7 i 4 5 k 2 ✓ ◆ 12(4) 3 0.081 1+ 61 7 12 6 7 9600 = P 6 7 0.081 4 5 12 " # 1 (1.00675) 48 9600 = P 0.00675 9600 = 40.883226P P = $234.81

A Special Number e. The number e ⇡ 2.7183 is often used as the base for applications of exponential functions in biology, physics and finance. Many (but not all) of these applications involve the use of a continuous growth or decay model. Called the natural exponential function, such a model has the form

Bradshaw - Math 188

Chapter 4 Notes 8 yo = beginning amount > > > < y(t) = amount at any time t y(t) = yo ert where > r = growth or decay constant > > : t = time

Example 4.5 : The radioactive isotope Plutonium-239 decays according to the model y(t) = yo e If you start with 500 kg of Plutonium, how much will remain after 1000 years?

44

0.00002876t .

Solution: Substituting t = 1000 and yo = 500 into the equation, we have y(t) = 500e

0.00002876(1000)

= 485.8 kg.

Section 4.3 Logarithmic Functions A logarithmic function with base b has the form f ( x ) = logb x where b is a constant, b > 0 and b 6= 1. A logarithmic function is the inverse of an exponential function. Please read the previous sentence at least three times. Using mathematical notation, this means that these two equations are equivalent. bx = y

()

logb y = x

()

log2 8 = 3

6 = 2 say the same thing but with the inverse operation 3 (multiplication versus division), the following two equations are equivalent. In the same way that the 2 ⇥ 3 = 6 and

23 = 8

Example 4.6 : Find the inverse function of f ( x ) = 3x . Solution: Finding the inverse function requires switching the variables and solving for y. This gives the the following. y = 3x x = 3y y = log3 x Therefore, f

1 (x)

= log3 x.

Example 4.7 : Write the exponent rules a0 = 1 and a1 = a in terms of logarithms. Solution: Using we have the following. and

bx = y

()

logb y = x,

a0 = 1

()

loga 1 = 0

a1 = a

()

loga a = 1

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Chapter 4 Notes

45

The concept of exponential and logarithmic functions being inverses plays a key role in understanding logarithmic functions. In particular, two features are worth pointing out. • Since the graph of a exponential function has a horizontal asymptote, the graph of a logarithmic function has a vertical asymptote. • For every rule of exponents, such as a0 = 1, there is a corresponding rule of logarithms, such as loga 1 = 0.

Graphs of Logarithmic Functions Since a log function is the inverse of an exponential function, the graph of the log function can be graphed by looking at the reversed points of an exponential function. Example 4.8 : Sketch the graph of f ( x ) = log2 x. Solution: Since the inverse of f ( x ) = log2 x is f ( x ) = 2x , first look at the graph of 2x , shown as the dashed curve. Using this information, we can reverse the order of the coordinates of each point and sketch the graph of f ( x ) = log2 x. The logarithm (solid curve) is the reflection of the exponential (dashed curve) across the line y = x (dotted line). Notice that the logarithm has a vertical asymptote at x = 0 while the exponential has a horizontal asymptote at y = 0. 8 6 4 2 6

4

2

2

4

6

8

2 4

x 1/8 1/4 1/2 1 2 4 8

y 3 2 1 0 1 2 3

6 Figure 4.4: Graph of f ( x ) = log2 x (solid curve) on 0 < x  8.

Special Logarithmic Functions There are two special functions, the natural logarithm and the common logarithm, which are used in the standard applications of exponential and logarithmic functions. The natural logarithm function is the inverse of f ( x ) = e x and is written as f ( x ) = loge x

or more commonly,

f ( x ) = ln x

The common logarithm function is the inverse of f ( x ) = 10x and is written as f ( x ) = log10 x

or more commonly,

f ( x ) = log x

Using these special logarithms, we can now solve two new types of equations.

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Chapter 4 Notes

46

Example 4.9 : Solve 10x = 400. Solution: Since 102 = 100 and 103 = 1000, we expect that solution will be have a value between 2 and 3. Using the property that an exponential can be written as a logarithm, we have 10x = 400

=)

x = log10 500 = log 400

Using a calculator, this gives x ⇡ 2.602.

Example 4.10 : The population P(t) of bacteria in a colony is given by P(t) = 100e0.023t , where t is time in days. Determine when there are 200 bacteria in the colony. Solution: Starting with 100e0.023t = 2000 and dividing by 100 gives e0.023t = 2 Using the property that an exponential can be written as a logarithm, we have e0.023t = 2

=)

0.023t = loge 2 = ln 2

Dividing by 0.023 and using a calculator gives t=

ln 2 ⇡ 30.14 minutes 0.023

Section 4.4 Laws of Logarithms As mentioned in the previous section, the exponent rules can be rewritten as logarithm rules. Below is the full list.

Logarithm Rules 1. logb x = y () x = by

6. logb xy = logb x + logb y

2. blogb x = x

7. logb

3. logb

bx

=x

4. logb 1 = 0 5. logb b = 1

x = logb x y

logb y

8. logb x y = y logb x 9. logb x =

loga x loga b

The logarithm rules can be used to either expand or collect terms in a logarithmic expression. Example 4.11 : Expand the expression log5 25x3 y4 . Solution: According to the sixth rule of the list, terms multiplied together inside of a logarithm can be treated as separate logarithms added together. This gives ⇣ ⌘ log5 25x3 y4 = log5 25 + log5 x3 + log5 y4

= log5 52 + log5 x3 + log5 y4

According to the eighth rule in the list, exponents on the inside of a logarithm can be brought to the outside of the logarithm. This gives log5 52 + log5 x3 + log5 y4 = 2 log5 5 + 3 log5 x + 4 log5 y. Finally, the fifth rule on the the list gives log5 5 = 1 so the final result is ⇣ ⌘ log5 25x3 y4 = 2 + 3 log5 x + 4 log5 y.

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Chapter 4 Notes

47

Example 4.12 : Collect the following terms into a single logarithmic expression. 2 + 5 log2 y

log2 x

Solution: Using a similar set of steps as in the previous example, we have the following. 2 + 5 log2 y

log2 x = 2 log2 2 + 5 log2 y 2

= log2 2 + log2 y ✓ 5◆ 4y = log2 x

5

log2 x log2 x

Example 4.13 : Simplify the expression 2

log2 x .

Solution: The coefficient of the logarithm is exponent. Therefore we have

1 so this can be brought inside of the logarithm as an 2

log2 x

= 2log2 x

1

Now, use the second rule on the list to finish the simplification. 2

log2 x

= 2log2 x

1

=x

1

=

1 x

Section 4.5 Exponential and Logarithmic Equations The goal of this section is to solve equations. For most examples, I will use the natural logarithm (ln) but you could also use the common logarithm (log). Example 4.14 : Solve 5x

3

= 17.

Solution: Since these are not nice numbers, take the logarithm of both sides of the equation and then solve for x. 5x 3 = 17 h i ln 5x 3 = ln 17

(x

Example 4.15 : Solve log4 ( x

3) + log4 ( x

3) ln 5 = ln 17 ln 17 x 3= ln 5 ln 17 x = 3+ ln 5 2) = 2.

Solution: Start by combining the logarithms into a single expression. Next, convert the logarithm equation into an exponential equation and solve the result for x. log4 ( x

3) + log4 ( x

2) = 2

log4 [( x 3)( x 2)] = 2 h i log4 x2 5x + 6 = 2 x2

x2

5x + 6 = 42

5x

10 = 0 x=

5±

p 2

65

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Chapter 4 Notes

48

Note that checking the solution gives two results that work. Example 4.16 : Solve log( x + 3)

log(2x

1) =

2.

Solution: Start by combining the logarithms into a single expression. Next, convert the logarithm equation into an exponential equation and solve the result for x. log( x + 3)

log(2x 1) = 2 x+3 log = 2 2x 1 x+3 = 10 2 2x 1 100x + 300 = 2x 1 301 x= 98

Note that checking the solution fails. Therefore, there is no real solution. Example 4.17 : Solve x

2 = 3 ln x.

Solution: This problem involves the variable x both inside the logarithm and by itself. Problems of this type do not have an algebraic solution. Therefore, we can find an estimate of the solution by looking at the the graph of the function f ( x ) = x 2 3 ln x.The graph shows there are two solutions, one solution between x = 0 and x = 2, and the other near x = 8. Zooming-in with a calculator gives x ⇡ 0.6343 and x ⇡ 8.376. 20

15

10

5

2

Figure 4.5: Graph of f ( x ) = x

4

2

6

8

10

3 ln x using 0  x  10,

5  y  20

Applications The remainder of this section uses the rules of logarithms and exponentials along the techniques for solving equations to answer questions about real-world situations. The financial and radioactive decay models were introduced in the previous sections. They are used again here along with some new models.

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Chapter 4 Notes

49

Example 4.18 : [Compound Interest] You make a single deposit of $2500 into a bank account. The account earns 6.6%, compounded monthly. How long will it take the value of the account to double? Solution: This example uses the compound interest formula introduced in the first section. From the statement of the problem, we have P = 2500, k = 12, and r = 0.066. Since the question asks for the time it takes the money to double, we have A = 5000. Substituting this into the formula, we have the following. ⇣ r ⌘kt A = P 1+ k ✓ ◆ 0.066 12t 5000 = 2500 1 + 12 2 = 1.005512t

ln 2 = 12t ln 1.0055 ln 2 t= ⇡ 11 months 12 ln 1.0055 Example 4.19 : [Annuity Formula] You have started your first job and decide to put $200 a month into an annuity. The annuity earns 7.2% interest, compounded monthly. How long (in months and years) will it take for the account to be worth $1,000,000? Solution: The annuity formula was introduced at the beginning of this chapter. 8 > > A = total amount in the account > 2⇣ 3 > r ⌘kt > > P = amount of your deposits < 1+ 1 6 7 k r = annual interest rate A = P4 where 5 r > > > t = time in years > k > > : k = number of periods per year Using A = 1, 000, 000, r = 0.072, we have the following. 2⇣

6 A = P4

1+

2✓

6 6 1, 000, 000 = 200 6 4

r ⌘kt k r k

1

3 7 5

◆ 0.072 12t 1+ 12 0.072 12

1.00612t 1 0.006 30 = 1.00612t 1

5000 =

31 = 1.00612t ln 31 = 12t ln 1.006 ln 31 t= ⇡ 48 years 12 ln 1.006

3

17 7 7 5

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Chapter 4 Notes

50

Example 4.20 : [Newton’s Law of Cooling] Newton’s Law of Cooling describes the temperature of an object as it cools (or heats). 8 T = the temperature of the object > > > > > > < T0 = initial temperature of the object T (t) = M + ( T0

M) e

kt

where

M = temperature of the surrounding area > > > t = time > > > : k = a constant

A cup of coffee has a temperature of 200 F and is placed in a room with temperature 65 F. After 4 minutes, the temperature of the coffee is 112 F. Write the equation for the temperature model of the coffee and find the temperature of the coffee after an additional 4 minutes. Solution: We start by substituting the known information into the equation. This gives the following. T (t) = M + ( T0

M) e

T (t) = 65 + (200 T (t) = 65 + 135e

65)e

kt kt

kt

We also know that at time t = 4 minutes, the temperature is 112 . Substituting these values allows solving for k. T (t) = 65 + 135e kt 112 = 65 + 135e 4k 47 = e 4k 135 ✓ ◆ 1 47 k= ln ⇡ 0.264 4 135

Therefore, the model for the temperature of the coffee at any time is given by T (t) = 65 + 135e

0.264t

After eight minute, the temperature of the coffee is T (8) = 65 + 135e

0.264(8)

= 81.3 F.