1. Chapter 7 - Exponential and Logarithmic Functions
Algebra 2

2. Table of Contents 7.1 – Exponential Functions, Growth and Decay
7.2 - Inverses of Relations and Functions
7.3 - Logarithmic Functions
7.4 - Properties of Logarithms
7.5 - Exponential and Logarithmic Equations and Inequalities
The Natural Base, e
7.7 - Transforming Exponential and Logarithmic Functions

3. 7.1 - Exponential Functions, Growth, and Decay
Algebra 2

4. The parent exponential function is f(x) = bx,
7.1
Algebra 2 (bell work)
The parent exponential function is f(x) = bx,
where the base b is a constant and the exponent x is the independent variable.

5. The graph of the parent function f(x) = 2x is shown.
7.1
Just Read
The graph of the parent function
f(x) = 2x is shown.
The domain is all real numbers and the range is {y|y > 0}.

6. 7.1
Notice as the x-values decrease, the graph of the function gets closer and closer to the x-axis.
The function never reaches the x-axis because the value of 2x cannot be zero. In this case, the x-axis is an asymptote.
An asymptote is a line that a graphed function approaches as the value of x gets very large or very small. as·ymp·tote
b > 1, exponential growth
0 < b < 1, exponential decay

7. 7.1
Example 1
Graphing Exponential Functions
Tell whether the function shows growth or decay. Then graph.
Step 1 Find the value of the base.
The base , ,is less than 1. This is an exponential decay function.
Step 2 Graph the function by using a table of values. x 2 4 6 8 10 12
f(x)
5.6
3.2
1.8
1.0
0.6
0.3

8. 7.1
Tell whether the function shows growth or decay. Then graph.
g(x) = 100(1.05)x
Step 1 Find the value of the base.
The base, 1.05, is greater than 1. This is an exponential growth function.
g(x) = 100(1.05)x

9. 7.1
Tell whether the function p(x) = 5(1.2x) shows growth or decay. Then graph.
Optional
Step 1 Find the value of the base.
The base , 1.2, is greater than 1. This is an exponential growth function.
p(x) = 5(1.2 x)
Step 2 Graph the function by using a table of values. x
–12 –8 –4 4 8 10
f(x)
0.56
1.2
2.4 5
10.4
21.5
30.9

10. Math Joke Q: How did you know that your dentist studied algebra?
7.1
Math Joke
Q: How did you know that your dentist studied algebra?
A: She said all that candy gave me exponential decay.

11. Read Student to Student pg. 491
7.1
Algebra 2 (bell work)
Day 2
You can model growth or decay by a constant percent increase or decrease with the following formula:
In the formula, the base of the exponential expression, 1 + r, is called the growth factor. Similarly, 1 – r is the decay factor.
Read Student to Student pg. 491

12. Clara invests $5000 in an account that pays 6.25% interest per year.
7.1
Example 2/3
Application
Clara invests $5000 in an account that pays 6.25% interest per year.
After how many years will her investment be worth $10,000?
Step 1 Write a function to model the growth in value of her investment.
f(t) = a(1 + r)t
f(t) = 5000( )t
f(t) = 5000(1.0625)t
Step 2 When graphing exponential functions in an appropriate domain, you may need to adjust the range a few times to show the key points of the function.
Step 3 Use the graph to predict when the value of the investment will reach $10,000.
Use the feature to find the t-value where f(t) ≈ 10,000.
** Can also use 2nd , Table to find it**

13. 7-1
Table Feature
The function value is approximately 10,000 when t ≈ The investment will be worth $10,000 about years after it was purchased.

14. 7.1
A city population, which was initially 15,500, has been dropping 3% a year.
Write an exponential function and graph the function. Use the graph to predict when the population will drop below 8000.
f(t) = a(1 – r)t
f(t) = 15,500(1 – 0.03)t
f(t) = 15,500(0.97)t
Graph the function. Use to find when the population will fall below 8000.
It will take about 22 years for the population to fall below 8000.

15. 7.1
In 1981, the Australian humpback whale population was 350 and increased at a rate of 14% each year since then.
Write a function to model population growth. Use a graph to predict when the population will reach 20,000.
P(t) = a(1 + r)t
P(t) = 350( )t
P(t) = 350(1.14)t
Graph the function. Use to find when the population will reach 20,000.
It will take about 31 years for the population to reach 20,000.

16. Use the graph to predict when the value will fall below $100.
7.1
A motor scooter purchased for $1000 depreciates at an annual rate of 15%. Write an exponential function and graph the function.
Use the graph to predict when the value will fall below $100.
f(t) = a(1 – r)t
f(t) = 1000(1 – 0.15)t
f(t) = 1000(0.85)t
Graph the function. Use to find when the value will fall below 100.
It will take about 14.2 years for the value to fall below 100.

17. Also mark the coordinates of the y-int
HW pg. 493
7.1-
Day 1: 2-4, 42-49
Day 2: 5, 6, 10-15, 29
On all graphs, minimum 5 points in a table. Pick x values on each side of the y-axis
Also mark the coordinates of the y-int

18. 7.2 - Inverses of Relations and Functions
Algebra 2

19. 7.2
Algebra 2 (bell work)
Summarize
To graph the inverse relation, you can reflect each point across the line y = x.
This is equivalent to switching the x- and y-values in each ordered pair of the relation.
Or it is equivalent to switching the domain and range

20. 7.2
Graph the relation and connect the points.
Then graph the inverse. Identify the domain and range of each relation.
Just Watch ● x 1 5 8 y 2 6 9 ● ● ●
Graph each ordered pair and connect them.
Switch the x- and y-values in each ordered pair. x 2 5 6 9 y 1 8

21. 1 5 8 2 6 9 • • • • • Domain:{x|0 ≤ x ≤ 8} Range :{y|2 ≤ x ≤ 9}
7.2
Just Watch
Reflect each point across y = x, and connect them.
Make sure the points match those in the table. • • • • •
Domain:{x|0 ≤ x ≤ 8} Range :{y|2 ≤ x ≤ 9}
Domain:{x|2 ≤ x ≤ 9} Range :{y|0 ≤ x ≤ 8} • • •
Blue Line
Red Line x 1 5 8 y 2 6 9 x 2 5 6 9 y 1 8

22. 7.2
Example 1
Graphing Inverse Relations
# 2 pg. 501 X 1 2 3 4 Y 8
Graph the relation. Graph the Inverse. Identify the domain and range of each relation

23. Functions that undo each other are inverse functions.
7.2
When the relation is also a function, you can write the inverse of the function f(x) as f –1(x). This notation does not indicate a reciprocal.
Functions that undo each other are inverse functions.
To find the inverse function, use the inverse operation. In the example above, 6 is added to x in f(x), so 6 is subtracted to find f–1(x).

24. Steps for solving for the inverse 1. Change f(x) into y =
7.2
Example 2
Writing Inverse Functions by Using Inverse Operations
Use inverse operations to write the inverse of f(x) = x – if possible. 1 2
Steps for solving for the inverse
1. Change f(x) into y =
2. Switch x and y
3. Solve for y again
4. Replace y with f –1(x)
f(x) = x – 1 2 1 2
f–1(x) = x +

25. Check Use the input x = 1 in f(x).
7.2
Check Use the input x = 1 in f(x).
Optional
f(x) = x – 1 2
f(1) = 1 – 1 2
Substitute 1 for x. = 1 2
Substitute the result into f–1(x) 1 2
f–1(x) = x + 1 2
f–1( ) = 1 2
Substitute for x.
= 1
The inverse function does undo the original function. 

26. Math Joke Q: How did the chicken find the inverse?
7.2
Math Joke
Q: How did the chicken find the inverse?
A: It reflected the function across y = eggs

27. Use inverse operations to write the inverse of f(x) = x 3
7.2
Use inverse operations to write the inverse of f(x) = x 3 x 3
f(x) =
f–1(x) = 3x

28. Check Use the input x = 1 in f(x).
7.2
Check Use the input x = 1 in f(x).
Optional x 3
f(x) = 1 3
f(1) =
Substitute 1 for x. = 1 3
Substitute the result into f–1(x)
f–1(x) = 3x
f–1( ) = 3( ) 1 3 1 3
Substitute for x.
= 1
The inverse function does undo the original function. 

29. Use inverse operations to write the inverse of f(x) = x +
7.2 2 3
Use inverse operations to write the inverse of f(x) = x +
f(x) = x + 2 3 2 3
f–1(x) = x –

30. Check Use the input x = 1 in f(x).
7.2
Check Use the input x = 1 in f(x).
Optional
f(x) = x + 2 3
f(1) = 1 + 2 3
Substitute 1 for x. = 5 3
Substitute the result into f–1(x) 2 3
f–1(x) = x – 2 3
f–1( ) = – 5 5 3
Substitute for x.
= 1
The inverse function does undo the original function. 

31. 7.2
Example 3
Writing Inverses of Multi-Step Functions
Use inverse operations to write the inverse of f(x) = 3(x – 7).
f(x) = 3(x – 7) 1 3
f–1(x) = x + 7
Check Use a sample input.  1 3
f–1(6) =(6) = 2 + 7= 9
f(9) = 3(9 – 7) = 3(2) = 6

32. 7.2
Use inverse operations to write the inverse of f(x) = 5x – 7.
f(x) = 5x – 7
f–1(x) =
x + 7 5
Check Use a sample input.
f–1(3) = = = 2 
f(2) = 5(2) – 7 = 3 10 5
3 + 7

33. Graph f(x) = – x – 5 . Then write the inverse and graph.
7.2
Example 4
Writing and Graphing Inverse Functions
Day 2
Graph f(x) = – x – Then write the inverse and graph. 1 2 1 2
y = – x – 5 1 2
x = – y – 5
x + 5 = – y 1 2
–2x – 10 = y
y = –2(x + 5)
f–1(x) = –2(x + 5)
f–1(x) = –2x – 10

34. 7.2 1
f(x) = – x – 5 2
f–1(x) = -2x - 10

35. Graph f(x) = x + 2. Then write the inverse and graph.
7.2
Graph f(x) = x Then write the inverse and graph. 2 3 2 3
y = x + 2 2 3
x = y + 2
x – 2 = y 2 3
3x – 6 = 2y
x – 3 = y 3 2
f–1(x) =
x – 3 3 2

36. 7.2
f–1(x) =
x – 3 3 2 2
Graph f(x) = x + 2 3

37. The list price of the CD is $14.
7.2
Example 5
Retail Application
Juan buys a CD online for 20% off the list price.
He has to pay $2.50 for shipping.
The total charge is $ What is the list price of the CD?
c = 0.80L
Step 1 Write an equation for the total charge as a function of the list price.
c – 2.50 = 0.80L
Step 2 Find the inverse function that models list price as a function of the change.
c – 2.50 = L
0.80
L =
13.70 – 2.50
0.80
Step 3 Evaluate the inverse function for c = $13.70.
= 14
The list price of the CD is $14.

38. 36 ounces of water should be added.
7.2
Optional
To make tea, use teaspoon of tea per ounce of water plus a teaspoon for the pot.
Use the inverse to find the number of ounces of water needed if 7 teaspoons of tea are used. 1 6 1 6
t = z + 1
Step 1 Write an equation for the number of ounces of water needed. 1 6
t – 1 = z
Step 2 Find the inverse function that models ounces as a function of tea.
6t – 6 = z
Step 3 Evaluate the inverse function for
t = 7.
z = 6(7) – 6 = 36
36 ounces of water should be added.

39. HW pg. 501 7.2- Day 1: 3-13, 61-65 (Odd) Day 2: 14, 15, 17, 27
7-2
HW pg. 501
7.2-
Day 1: 3-13, (Odd)
Day 2: 14, 15, 17, 27
Ch: 29-31, 34-40, 47

40. 7.3 - Logarithmic Functions
Algebra 2

41. 7.3
Algebra 2 (bell work)
You can write an exponential equation as a logarithmic equation and vice versa.
Read logb a= x, as “the log base b of a is x.” Notice that the log is the exponent.
Reading Math

42. ab = c 35 = 243 log3243 = 5 104 = 10,000 Exponential Equation
7.3
Example 1
Converting from Exponential to Logarithmic form
Write each exponential equation in logarithmic form.
Exponential Equation
Logarithmic Form
35 = 243
104 = 10,000
ab = c
log3243 = 5 1 2
log255 =
log1010,000 = 4 1 6
log = –1
logac =b

43. 7.3
Example 2
Converting from Logarithmic to Exponential Form
Write each logarithmic form in exponential equation.
Logarithmic Form
Exponential Equation
log99 = 1
log = 9
logb1 = 0
91 = 9
29 = 512 1 16
4–2 =
b0 = 1

44. A logarithm with base 10 is called a common logarithm.
7.3
A logarithm with base 10 is called a common logarithm.
If no base is written for a logarithm, the base is assumed to be 10.
For example, log 5 = log105.

45. 7.3
Math Joke
Q: Why are you drumming on your algebra book with two big sticks?
A: Because we’re studying log rhythms

46. 7.3
Example 3
Evaluating Logarithms by Using Mental Math
Evaluate by using mental math. 1 5
log 0.01
log5 125
log5
10? = 0.01
5? = 125
5? = 1 5 -2
53 = 125
10 = 0.01
log5125 = 3 = 1 5
log 0.01 = –2
log = –1 1 5

47. 7.3
Algebra 2 (bell work)
Day 2
Because logarithms are the inverses of exponents, the inverse of an exponential function, such as y = 2x, is a logarithmic function, such as y = log2x.
You may notice that the domain and range of each function are switched.
The domain of y = 2x is all real numbers (R), and the range is {y|y > 0}.
The domain of y = log2x is {x|x > 0}, and the range is all real numbers (R).

48. f(x) = 1.25x 7.3 Example 4 Graphing Logarithmic Functions On Calc
Use the x-values {–2, –1, 0, 1, 2}.
Graph the function and its inverse. Describe the domain and range of the inverse function.
f(x) = 1.25x
Graph f(x) = 1.25x by using a table of values. 1
f(x) = 1.25x 2 –1 –2 x
0.64
0.8
1.25
1.5625

49. 7.3
To graph the inverse, f–1(x) = log1.25x, by using a table of values. 2 1 –1 –2
f–1(x) = log1.25x
1.5625
1.25
0.8
0.64 x
The domain of f–1(x) is {x|x > 0}, and the range is R.

50. 7.3
Use the x-values {–2, –1, 0, 1, 2}. Graph the function and its inverse. Describe the domain and range of the inverse function.
f(x) = x 1 2
Graph f(x) = x by using a table of values. 1 2 x –2 –1 1 2
f(x) =( ) x 4 1 2 4

51. 7.3 1 2
To graph the inverse, f–1(x) = log x, by using a table of values. x 4 2 1
f –1(x) =log x –2 –1 1 2 4
The domain of f–1(x) is {x|x > 0}, and the range is R.

52. Lemon juice has the pH of about 2.2. Milk has the pH of about 6.6.
7.3
Example 5
Environmental application
The table lists the hydrogen ion concentrations for a number of food items. Find the pH of each.
Substance
H+ conc. (mol/L)
Milk
Tomatoes
Lemon juice
0.0063
pH = –log[H+ ]
Lemon juice has the pH of about 2.2.
Milk has the pH of about 6.6.
pH = –log( )
pH = –log(0.0063)
Tomatoes have the pH of about 4.5.
pH = –log( )

53. 7-3
HW pg. 509
7.3-
Day 1: 1-13, 25-28, 33, 38
Day 2: 14-16, 31, 48-52
Ch: 32, 34, 35-37, 44-47
On graphs need 5 coordinates listed in a table for each function f(x) and f-1(x)

54. 7.4 - Properties of Logarithms
Algebra 2

55. Think: logj + loga + logm = logjam
7.4
Algebra 2 (bell work)
Remember that to multiply powers with the same base, you add exponents.
Pg. 512
Helpful Hint
Think: logj + loga + logm = logjam

56. log64 + log69 log5625 + log525 7.4 Example 1 Adding Logarithms
Express as a single logarithm. Simplify, if possible.
log64 + log69
log log525
log6 (4  9)
log5 (625 • 25)
log5 15,625 =6
log6 36 = 2 6 2 1 3
log = -1 –1

57. 7.4
Remember that to divide powers with the same base, you subtract exponents
Pg. 513
The property above can also be used in reverse.
Just as a5b3 cannot be simplified, logarithms must have the same base to be simplified.
Caution

58. log5100 – log5 4 log749 – log7 7 7.4 Example 2 Subtracting Logarithms
Express as a single logarithm. Simplify, if possible.
log5100 – log5 4
log749 – log7 7
log5(100 ÷ 4)
log7(49 ÷ 7)
Log7= 1
Log525 =2 2 1

59. Math Joke Teacher: What are some properties of logs?
7.4
Math Joke
Teacher: What are some properties of logs?
Student: They’re round, they have bark, and they come from trees

60. 7.4
Because you can multiply logarithms, you can also take powers of logarithms.
Pg. 513

61. 7.4
Example 3
Simplifying Logarithms with Exponents
Express as a product. Simplify, if possible.
A. log2326
B. log8420
20log84
6log232
20( ) = 40 3 2
6(5) = 30
Because 25 = 32, log232 = 5.
Because = 4,
log84 = 2 3

62. a. log104 b. log5252 7.4 Express as a product. Simplify, if possibly.
4(1) = 4
2(2) = 4
Because 101 = 10, log 10 = 1.
Because 52 = 25, log525 = 2.

63. 7.4
Day 2
Exponential and logarithmic operations undo each other since they are inverse operations.
Pg. 514
Pg. 514

64. a. log3311 b. log381 c. 5log 10 Simplify each expression.
7.4
Example 4
Recognizing Inverses
Simplify each expression.
a. log3311
b. log381
c. 5log 10 5
log33  3  3  3
log3311
5log510
log334 10 11 4

65. a. Simplify log100.9 b. Simplify 2log2(8x) 2log2(8x) log 100.9 8x 0.9
7.4
a. Simplify log100.9
b. Simplify 2log2(8x)
2log2(8x)
log 100.9 8x
0.9

66. Evaluate log328 log8 log28 log32 log232 7.4 Example 5
Changing the Base of a Logarithm
Evaluate log328
Method 1 Change to base 10
Method 2 Change to base 2, because both 32 and 8 are powers of 2.
log328 =
log8
log32
log328 =
log28
log232 = 3 5
0.903
1.51 ≈
= 0.6
≈ 0.6

67. Evaluate log927 log27 log327 log9 log39 7.4
Method 1 Change to base 10.
Method 2 Change to base 3, because both 27 and 9 are powers of 3.
log927 =
log27
log9
log927 =
log327
log39 = 3 2
1.431
0.954 ≈
= 1.5
≈ 1.5

68. Evaluate log816. log16 log416 log8 log48 7.4 Optional
Method 1 Change to base 10.
Method 2 Change to base 4, because both 16 and 8 are powers of 2.
Log816 =
log16
log8
log816 =
log416
log48 = 2
1.5
1.204
0.903 ≈
= 1.3
≈ 1.3

69. HW pg.516 7.4- Day 1: 1-10, 21-31 (Odd), 81-95 (Odd)
Ch: 19, 35-39, 46-50, 54-65, 69-80

70. 7.5 - Exponential and Logarithmic Equations and Inequalities
Algebra 2

71. To solve exponential equations:
7.5
Algebra 2 (bell work)
An exponential equation is an equation containing one or more expressions that have a variable as an exponent.
To solve exponential equations:
Try writing them so that the bases are all the same.
Take the logarithm of both sides.

72. 98 – x = 27x – 3 7.5 Example 1 Solving Exponential Equations
Solve and check.
98 – x = 27x – 3
Rewrite each side with the same base; 9 and 27 are powers of 3.
(32)8 – x = (33)x – 3
To raise a power to a power, multiply exponents.
316 – 2x = 33x – 9
16 – 2x = 3x – 9
Bases are the same, so the exponents must be equal.
x = 5
Solve for x.

73. 4x – 1 = 5 log5 log4 Log5 log4 7.5 Solve and check.
5 is not a power of 4, so take the log of both sides.
log 4x – 1 = log 5
Apply the Power Property of Logarithms.
(x – 1)log 4 = log 5
x –1 =
log5
log4
Divide both sides by log 4.
x = ≈ 2.161
Log5
log4
Check Use a calculator.
The solution is x ≈

74. 7–x = 21 log21 log7 log21 log7 7.5 Solve and check.
21 is not a power of 7, so take the log of both sides.
log 7–x = log 21
(–x)log 7 = log 21
Apply the Power Property of Logarithms.
–x =
log21
log7
Divide both sides by log 7.
x = – ≈ –1.565
log21
log7

75. 23x = 15 log15 log2 7.5 Solve and check. Optional
15 is not a power of 2, so take the log of both sides.
log23x = log15
Apply the Power Property of Logarithms.
(3x)log 2 = log15
3x =
log15
log2
Divide both sides by log 2, then divide both sides by 3.
x ≈ 1.302

76. Check In 20 hours, there will be 220 bacteria.
7.5
Example 2
Application
Suppose a bacteria culture doubles in size every hour. How many hours will it take for the number of bacteria to exceed 1,000,000?
At hour 0, there is one bacterium, or 20 bacteria. At hour one, there are two bacteria, or 21 bacteria, and so on. So, at hour n there will be 2n bacteria.
Write 1,000,000 in scientific annotation.
Solve 2n > 106
log 2n > log 106
Take the log of both sides.
Use the Power of Logarithms.
nlog 2 > log 106
nlog 2 > 6
log 106 is 6. 6
log 2
n >
Divide both sides by log 2. 6
0.301
n >
Evaluate by using a calculator.
n > ≈ 19.94
Round up to the next whole number.
It will take about 20 hours for the number of bacteria to exceed 1,000,000.
Check In 20 hours, there will be 220 bacteria.
220 = 1,048,576 bacteria.

77. Get a single log on each side of the equation
7.5
Day 2
A logarithmic equation is an equation with a logarithmic expression that contains a variable.
You can solve logarithmic equations by using the properties of logarithms.
Get a single log on each side of the equation
Use the inverse property or the property below to solve (make each side exponential)

78. log6(2x – 1) = –1 1 6 7 12 7.5 Example 3 Solving Logarithmic Equations
Solve
log6(2x – 1) = –1
6 log6 (2x –1) = 6–1
Use 6 as the base for both sides.
2x – 1 = 1 6
Use inverse properties to remove 6 to the log base 6. 7 12
x =
Simplify.

79. log4100 – log4(x + 1) = 1 100 x + 1 100 x + 1 7.5 Solve log4( ) = 1
Write as a quotient.
4log = 41
100
x + 1
( )
Use 4 as the base for both sides.
= 4
100
x + 1
Use inverse properties on the left side.
x = 24

80. Math Joke Teacher: Did you get the answer?
7.5
Math Joke
Teacher: Did you get the answer?
Student: Well, I got closer and closer to it, but I guess it was just an asymptote

81. log5x 4 = 8 7.5 Solve 4log5x = 8 Power Property of Logarithms.
Divide both sides by 4 to isolate log5x.
x = 52
Definition of a logarithm.
x = 25

82. log12x + log12(x + 1) = 1 7.5 Solve log12 x(x + 1) = 1
Product Property of Logarithms.
log12x(x +1)
= 121
Exponential form.
x(x + 1) = 12
Use the inverse properties.
x2 + x – 12 = 0
Multiply and collect terms.
(x – 3)(x + 4) = 0
Factor.
x – 3 = 0 or x + 4 = 0
Set each of the factors equal to zero.
Solve.
x = 3 or x = –4

83. 7.5
Check Check both solutions in the original equation.
log12x + log12(x +1) = 1
log12x + log12(x +1) = 1
log123 + log12(3 + 1) 1
log12( –4) + log12(–4 +1) 1
log123 + log
log12( –4) is undefined.
log
1 1 
The solution is x = 3.

84. 3 = log 8 + 3log x 2log x – log 4 = 0 x 4 x 4 7.5 Solve
Optional
3 = log 8 + 3log x
2log x – log 4 = 0
3 = log 8 + 3log x
2log( ) = 0 x 4
3 = log 8 + log x3
3 = log (8x3)
2(10log ) = 100 x 4
103 = 10log (8x3)
2 ( ) = 1 x 4
1000 = 8x3
125 = x3
x = 2
5 = x

85. HW pg. 526 7.5- Day 1: 2-8, 21-25 (Odd), 55-61 (Odd)
Ch: 40-47, 51-53

86. 7.6 - The Natural Base, e
Algebra 2

87. 7.6
Algebra 2 (bell work)
Do Not Copy
As n gets very large, interest is continuously compounded. Examine the graph of f(n)= ( )n.
The function has a horizontal asymptote.
As n becomes infinitely large, the value of the function approaches approximately ….
This number is called e. Like , the constant e is an irrational number. n 1

88. Do Not Copy The domain of f(x) = ex is all real numbers.
7.6
Do Not Copy
Exponential functions with e as a base have the same properties as the functions you have studied.
The graph of f(x) = ex is like other graphs of exponential functions, such as f(x) = 3x.
The domain of f(x) = ex is all real numbers.
The range is {y|y > 0}.

89. Math Joke Kurt: What are you studying? Amy: e!
7.6
Math Joke
Kurt: What are you studying?
Amy: e!
Kurt: Natural Logarithms?
Amy: No, I mean “eee!” There’s a wasp on your leg!

90. Graph f(x) = ex–2 + 1. Make a table. Round to nearest tenth
7.6
Example 1
Graphing Exponential Functions
Graph f(x) = ex–2 + 1.
Make a table.
Round to nearest tenth
Give me 5 Points (always if you use a calc)
Graph x –2 –1 1 2 3 4
f(x) = ex–2 + 1
1.0
1.1
1.4
3.7
8.4
Hint: You can graph the function, and use the table feature on your calculator
Or, Hit [2nd] [Enter] and just replace the x value

91. Give me 5 Points (always if you use a calc)
7.6
Graph f(x) = ex – 3.
Make a table.
Round to nearest tenth
Give me 5 Points (always if you use a calc)
Graph x –4 –3 –2 –1 1 2
f(x) = ex – 3
–2.9
–2.7
–0.3
4.4
Hint: You can graph the function, and use the table feature on your calculator
Or, Hit [2nd] [Enter] and just replace the x value

92. 7.6
A logarithm with a base of e is called a natural logarithm and is abbreviated as “ln” (rather than as loge).
Natural logarithms have the same properties as log base 10 and logarithms with other bases.
Pg. 532
Copy Picture
The natural logarithmic function f(x) = ln x is the inverse of the natural exponential function f(x) = ex.

93. The domain of f(x) = ln x is {x|x > 0}.
7.6
The domain of f(x) = ln x is {x|x > 0}.
Pg. 532
The range of f(x) = ln x is all real numbers.
All of the properties of logarithms from Lesson 7-4 also apply to natural logarithms.

94. A. ln e0.15t B. e3ln(x +1) C. ln e2x + ln ex Simplify
7.6
Example 2
Simplifying Expressions with e or ln
Simplify
A. ln e0.15t
B. e3ln(x +1)
e3ln(x +1) = (x + 1)3
ln e0.15t = 0.15t
C. ln e2x + ln ex
ln e2x + ln ex = 2x + x = 3x

95. a. ln e3.2 b. e2lnx c. ln ex +4y Simplify e2lnx = x2 ln e3.2 = 3.2
7.6
Simplify
Optional
a. ln e3.2
b. e2lnx
e2lnx = x2
ln e3.2 = 3.2
c. ln ex +4y
ln ex + 4y = x + 4y

96. The formula for continuously compounded interest is A = Pert
7.6
The formula for continuously compounded interest is A = Pert
-A is the total amount,
-P is the principal
-r is the annual interest rate
-t is the time in years.

97. Use the ex key on a calculator. The total amount is $4,083.08.
7.6
What is the total amount for an investment of $500 invested at 5.25% for 40 years and compounded continuously?
A = Pert
A = 500e0.0525(40)
A ≈
Use the ex key on a calculator.
The total amount is $4,

98. Use the ex key on a calculator. The total amount is $132.31.
7.6
What is the total amount for an investment of $100 invested at 3.5% for 8 years and compounded continuously?
A = Pert
A = 100e0.035(8)
A ≈
Use the ex key on a calculator.
The total amount is $

99. Natural decay is modeled by the function below.
7-6
Day 2
The half-life of a substance is the time it takes for half of the substance to breakdown or convert to another substance during the process of decay.
Natural decay is modeled by the function below.

100. Plutonium-239 (Pu-239) has a half-life of 24,110 years.
7.6
Example 4
Application
Plutonium-239 (Pu-239) has a half-life of 24,110 years.
How long does it take for a 1 g sample of Pu-239 to decay to 0.1 g?
Step 1 Find the decay constant for plutonium-239.
*Always solve for k (the rate) first*
N(t) = N0e–kt
Substitute 1 for N0 ,24,110 for t, and for N(t) because half of the initial quantity will remain. 1 2
= 1e–k(24,110) 1 2

101. Simplify and take ln of both sides.
7.6
ln = ln e–24,110k 1 2
Simplify and take ln of both sides.
Write as 2 –1, and simplify the right side. 1 2
ln 2–1 = –24,110k
ln2–1 = –1ln 2 = –ln 2
–ln 2 = –24,110k
ln2
24,110
k = ≈

102. Step 2 Write the decay function and solve for t.
7.6
Step 2 Write the decay function and solve for t.
Substitute for k.
N(t) = N0e– t
0.1 = 1e– t
Substitute 1 for N0 and 0.01 for N(t).
Take ln of both sides.
ln 0.1 = ln e– t
ln 0.1 = – t
Simplify.
ln 0.1
t = – ≈ 80,000
It takes approximately 80,000 years to decay.

103. Step 1 Find the decay constant for Chromium-51.
7.6
Determine how long it will take for 650 mg of a sample of chromium-51 which has a half-life of about 28 days to decay to 200 mg.
Step 1 Find the decay constant for Chromium-51.
N(t) = N0e–kt
Use the natural decay function. t.
= 1e–k(28) 1 2
Substitute 1 for N0 ,28 for t, and for N(t) because half of the initial quantity will remain. 1 2

104. Simplify and take ln of both sides.
7.6
ln = ln e–28k 1 2
Simplify and take ln of both sides.
Write as 2 –1 , and simplify the right side. 1 2
ln 2–1 = –28k
–ln 2 = –28k
ln 2 –1 = –1ln 2 = –ln 2.
k = ≈
ln 2 28

105. Step 2 Write the decay function and solve for t.
7.6
Step 2 Write the decay function and solve for t.
Substitute for k.
N(t) = N0e–0.0247t
Substitute 650 for N0 and 200 for N(t).
200 = 650e–0.0247t
ln = ln e–0.0247t
650
200
Take ln of both sides.
Simplify.
ln = –0.0247t
650
200
t = ≈ 47.7
650
200 ln –
It takes approximately 47.7 days to decay.

106. HW pg.534 7.6- Day 1: 3-11 (Odd), 17-21, 31-36 Day 2: 12, 22, 53-58
Ch: 25, 37-40
On all graphing problems need
Table with 5 points and graph

107. 7.7 - Transforming Exponential and Logarithmic Functions
Algebra 2

108. 7.7
Algebra 2 (bell work)
Copy Entire Table
Pg. 537

109. 7.7
Example 1
Translating Exponential Functions
Make a table of values, and graph g(x) = 2–x + 1.
Describe the asymptote.
Tell how the graph is transformed from the graph of the function f(x) = 2x. x –3 –2 –1 1 2
g(x) 9 5 3
1.5
1.25
The asymptote is y = 1, and the graph approaches this line as the value of x increases.
The transformation reflects the graph across the y-axis and moves the graph 1 unit up.
g(x) = 2–x + 1.

110. x –2 –1 1 2 f(x) 7.7 Make a table of values, and graph f(x) = 2x – 2.
Describe the asymptote.
Tell how the graph is transformed from the graph of the function f(x) = 2x. x –2 –1 1 2
f(x) 1 16 8 4 2
The asymptote is y = 0, and the graph approaches this line as the value of x decreases.
The transformation moves the graph 2 units right.
f(x) = 2x – 2.

111. 7.7
Example 2
Stretching, Compressing, and Reflecting Exponential Functions
Graph the function.
Find y-intercept and the asymptote.
Describe how the graph is transformed from the graph of its parent function.
A. g(x) = (1.5x) 2 3
Parent function: f(x) = 1.5x
The graph of g(x) is a vertical compression of the parent function f(x) 1.5x by a factor of 2 3
y-intercept: 2 3
asymptote: y = 0

112. 7.7
B. h(x) = e–x + 1
parent function: f(x) = ex
y-intercept: e
asymptote: y = 0
The graph of h(x) is a reflection of the parent function, f(x) = ex across the y-axis and a shift of unit to the right.
The range is {y|y > 0}.
h(x) = e–x + 1

113. Math Joke Lee: We need to stretch the exponential function
Sal: It looks stretched enough already!

114. 7.7
Copy Entire Table
Day 2
Pg. 538

115. 7.7
Example 3
Transforming Logarithmic Functions
Graph each logarithmic function.
Find the asymptote.
Describe how the graph is transformed from the graph of its parent function.
g(x) = 5 log x – 2
asymptote: x = 0
The graph of g(x) is a vertical stretch of the parent function
f(x) = log x by a factor of And a translation 2 units down.
g(x) = 5 log x – 2

116. 7.7
Graph each logarithmic function.
Find the asymptote.
Describe how the graph is transformed from the graph of its parent function.
h(x) = ln(–x + 2)
asymptote: x = 2
The graph of h(x) is a reflection of the parent function
f(x) = ln x across the y-axis and a shift of 2 units to the right. D:{x|x < 2}
h(x) = ln(–x + 2)

117. f(x) = 4x is reflected across both axes and moved 2 units down.
7.7
Example 4
Writing Transformed Functions
Write each transformed function.
f(x) = 4x is reflected across both axes and moved 2 units down. 4x
Begin with the parent function.
4–x
To reflect across the y-axis, replace x with –x.
–(4–x)
To reflect across the x-axis, multiply the function by –1.
g(x)= –(4–x) – 2
To translate 2 units down, subtract 2 from the function.

118. 7.7
f(x) = ln x is compressed horizontally by a factor of and moved 3 units left. 1 2
g(x) = ln2(x + 3)
When you write a transformed function, you may want to graph it as a check.

119. 7.7
Write the transformed function when f(x) = log x is translated 3 units left and stretched vertically by a factor of 2.
g(x) = 2 log(x + 3)
When you write a transformed function, you may want to graph it as a check.

120. Problem-Solving Application Skip
7.7
Example 5
Problem-Solving Application
Skip
The temperature in oF that milk must be kept at to last n days can be modeled by
T(n) = 75 – 16 ln n.
Describe how the model is transformed from f(n) = ln n.
Use the model to predict how long milk will last if kept at 34oF.
T(n) = 75 – 16 ln n
T(n) = –16 ln n + 75
The graph of f(n) = ln n is reflected across the x-axis, vertically stretched by a factor of 16, and translated 75 units up.
Find the number of days the milk will last at 34oF.
34 = –16ln n + 75
–41 = –16ln n
–41
–16
= ln n
e = n 41 16
n ≈ 13
The model predicts that the milk will last about 13 days. It is reasonable that milk would last 13 days if kept at 34oF.

121. 7.7
A group of students retake the written portion of a driver’s test after several months without reviewing the material.
A model used by psychologists describes retention of the material by the function a(t) = 85 – 15log(t + 1), where a is the average score at time t (in months).
Describe how the model is transformed from its parent function. When would the average score drop below 0. Is your answer reasonable?
a(t) = 85 – 15 log(t + 1)
a(t) = –15 log(t + 1) + 85
The graph of f(t) = ln n is reflected across the x-axis, vertically stretched by a factor of 15, and translated 85 units up and 1 unit left.
Find the time when the average score drops to 0.
0 = –15 log(t+1) + 85
–85 = –15 log(t + 1)
5.67 = log(t + 1)
= t + 1
It is unreasonable that scores would drop to zero 38,683, years after the students take the test without reviewing the material.
464,194 ≈ t
38,683 ≈ t

122. HW pg. 541 7.7- Day 1: 1-9 (Odd), 25-27 (No Graphs), 58-60
Ch: 31, 32