Lesson 9-1
Exponential Functions
Lesson 9-2
Logarithms and Logarithmic Functions
Lesson 9-3
Properties of Logarithms
Lesson 9-4
Common Logarithms
Lesson 9-5
Base e and Natural Logarithms
Lesson 9-6
Exponential Growth and Decay
Five-Minute Check (over Chapter 8)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Graph an Exponential Function
Key Concept: Exponential Growth and Decay
Example 2: Identify Exponential Growth and Decay
Example 3: Real-World Example: Write an Exponential Function
Key Concept: Property of Equality for Exponential Functions
Example 4: Solve Exponential Equations
Key Concept: Property of Inequality for Exponential Functions
Example 5: Solve Exponential Inequalities
• Graph exponential functions.
• Solve exponential equations and inequalities.
• exponential function
• exponential growth
• exponential decay
• exponential equation
• exponential inequality
2A.11 The student formulates equations and
inequalities based on exponential and logarithmic
functions, uses a variety of methods to solve them, and
analyzes the solutions in terms of the situation. (B) Use
the parent functions to investigate, describe, and
predict the effects of parameter changes on the
graphs of exponential and logarithmic functions,
describe limitations on the domains and ranges,
and examine asymptotic behavior. (F) Analyze a
situation modeled by an exponential function,
formulate an equation or inequality, and solve the
problem. Also addresses TEKS 2A.4(A), 2A.11(C),
2A.11(D), and 2A.11(E).
Graph an Exponential Function
Sketch the graph of y = 4x. State the function’s
domain and range.
Make a table of values. Connect the points to sketch a
smooth curve.
Graph an Exponential Function
Answer:
The domain is all real numbers, and the range is all
positive numbers.
Which is the graph of y = 3x?
0%
A.
B.
0%
C.
D.
A
B
0%
C
D
0%
D
D.
C
C.
B
B.
A
A.
Identify Exponential Growth and Decay
A. Determine whether
represents
exponential growth or decay.
Answer: The function represents exponential growth,
since the base,
, is greater than 1.
Identify Exponential Growth and Decay
B. Determine whether y = (0.7)x represents
exponential growth or decay.
Answer: The function represents exponential decay,
since the base, 0.7, is between 0 and 1.
A. What type of exponential function is represented
by y = (0.5)x?
A. exponential growth
B. exponential decay
C. both exponential
growth and
exponential decay
D. neither exponential
growth nor
exponential decay
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
B. What type of exponential function is represented
by
?
A. exponential growth
B. exponential decay
0%
C. both exponential
growth and
exponential decay
D. neither exponential
growth nor
exponential decay
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Write an Exponential Function
A. PHONES In 1995, there were an estimated
28,154,000 cellular telephone subscribers in the
United States. By 2005, this number had risen to
194,633,000.
Write an exponential function of the form y = abx that
could be used to model the number of cellular
telephone subscribers y in the U.S. Write the function
in terms of x, the number of years since 1995.
For 1995, the time x equals 0, and the initial number of
cellular telephone subscribers y is 28,154,000. Thus the
y-intercept, and the value of a, is 28,154,000.
For 2005, the time x equals 2005 – 1995 or 10, and the
number of cellular telephone subscribers is 194,633,000.
Write an Exponential Function
Substitute these values and the value of a into an
exponential function to approximate the value of b.
y = abx
Exponential function
194,633,000 = 28,154,000b10
Replace x with 10, y
with 194,633,000 and
a with 28,154,000.
6.91 ≈ b10
Divide each side by
28,154,000.
Take the 10th root of
each side.
Write an Exponential Function
To find the 10th root of 6.91, use selection
under the MATH menu on the TI-83/84 Plus.
Keystrokes:
10
MATH
5
6.91 ENTER
1.21324302926
Answer: An equation that models the number of cellular
telephone subscribers in the U.S. from 1995 to
2005 is y = 28,154,000(1.213)x.
Write an Exponential Function
B. Suppose the number of cellular subscribers
continues to increase at the same rate. Estimate the
number of U.S. subscribers in 2015.
For 2015, the time x equals 2015 – 1995 or 20.
y = 28,15400(1.213)x
Modeling equation
y = 28,15400(1.213)20
Replace x with 20.
y ≈ 1,338,924,608
Use a calculator.
Answer: The number of cell phone subscribers will be
about 1,339,000,000 in 2015.
A. HEALTH In 1991, 4.9% of Americans had diabetes.
By 2000, this percent had risen to 7.3%. Which
exponential equation could be used to model the
percentage of Americans with diabetes if the function
is in terms of x, the number of years since 1991?
A. y = 4.9(0.67)x
B. y = 7.3(0.67)
x
C. y = 7.3(1.05)x
0%
1.
2.
3.
4.
A
B
C
D
A
D. y = 4.9(1.05)x
B
C
D
B. Suppose the percent of Americans with diabetes
continues to increase at the same rate. What is a
good estimate for the percent of Americans with
diabetes in 2010?
A. 10.5
B. 11.4
C. 12.4
D. 13.0
0%
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Solve Exponential Equations
A. Solve 49n – 2 = 256.
49n – 2 = 256
Original equation
49n – 2 = 44
Rewrite 256 as 44 so each
side has the same base.
9n – 2 = 4
9n = 6
Property of Equality for
Exponential Functions
Add 2 to each side.
Divide each side by 9.
Answer: The solution is
.
Solve Exponential Equations
Check
49n – 2 = 256
Original equation
Substitute
Simplify.

Simplify.
for n.
Solve Exponential Equations
B. Solve 35x = 92x – 1.
35x = 92x – 1
Original equation
35x = 32(2x – 1)
Rewrite 9 as 32 so each
side has the same base.
5x = 2(2x – 1)
Property of Equality for
Exponential Functions
5x = 4x – 2
Distributive Property
x = –2
Subtract 4x from each side.
Answer: The solution is x = –2.
Solve Exponential Equations
Check
35x = 92x – 1
35(–2) = 92(–2) – 1
3–10 = 9–5
0.0000169 = 0.0000169 
Original equation
Substitute –2 for x.
Simplify.
Simplify.
A. What is the solution to the equation 23x + 1 = 32?
A. 1
B.
C.
D. 2
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. What is the solution to the equation 52x = 252x – 1?
A.
B.
C. 1
D. no solution
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Solve Exponential Inequalities
Solve
.
Original inequality
Rewrite
as
or 5–4.
Property of Inequality for
Exponential Functions.
Subtract 3 from each side.
Divide each side by –2.
Answer: The solution is
.
Solve Exponential Inequalities
Check:
Test a value of k less than
k = 0.
; for example,
Original inequality
Replace k with 0.
Simplify.

53 = 125
What is the solution to the inequality
?
A. k < 3
B. k > 3
C. k < 0
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 9-1)
Main Ideas and Vocabulary
Targeted TEKS
Key Concept: Logarithm with Base b
Example 1: Logarithmic to Exponential Form
Example 2: Exponential to Logarithmic Form
Example 3: Evaluate Logarithmic Expressions
Example 4: Solve a Logarithmic Equation
Key Concept: Logarithmic to Exponential Inequality
Example 5: Solve a Logarithmic Inequality
Key Concept: Property of Equality for Logarithmic Functions
Example 6: Solve Equations with Logarithms on Each Side
Key Concept: Property of Inequality for Logarithmic Functions
Example 7: Solve Inequalities with Logarithms on Each Side
• Evaluate logarithmic expressions.
• Solve logarithmic equations and inequalities.
• logarithm
• logarithmic function
• logarithmic equation
• logarithmic inequality
2A.11 The student formulates equations and
inequalities based on exponential and logarithmic
functions, uses a variety of methods to solve them, and
analyzes the solutions in terms of the situation.
(A) Develop the definition of logarithms by
exploring and describing the relationship between
exponential functions and their inverses.
(D) Determine solutions of exponential and
logarithmic equations using graphs, tables, and
algebraic methods. Also addresses TEKS 2A.4(A),
2A.11(C).
Logarithmic to Exponential Form
A. Write log3 9 = 2 in exponential form.
Answer: log3 9 = 2 → 9 = 32
Logarithmic to Exponential Form
B. Write
Answer:
in exponential form.
A. What is log2 8 = 3 written in exponential form?
A. 83 = 2
B. 23 = 8
C. 32 = 8
D. 28 = 3
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. What is
–2 written in exponential form?
A.
B.
C.
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Exponential to Logarithmic Form
A. Write 53 = 125 in logarithmic form.
Answer: 53 = 125 → log5 125 = 3
Exponential to Logarithmic Form
B. Write
Answer:
in logarithmic form.
A. What is 34 = 81 written in logarithmic form?
A. log3 81 = 4
B. log4 81 = 3
0%
C. log81 3 = 4
D. log3 4 = 81
1.
2.
3.
4.
A
B
C
D
A
B
C
D
B. What is
written in logarithmic form?
A.
B.
0%
C.
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Evaluate Logarithmic Expressions
Evaluate log3 243.
log3 243 = y
243 = 3y
35 = 3y
5 =y
Let the logarithm equal y.
Definition of logarithm
243 = 35
Property of Equality for
Exponential Functions
Answer: So, log3 243 = 5.
Which value is equivalent to log10 1000?
A.
0%
B. 3
C. 30
D. 10,000
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Solve a Logarithmic Equation
Solve
.
Original equation
Definition of logarithm
8 = 23
Power of a Power
Answer: n = 16
What is the solution to
?
A.
B. 3
C. 9
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Animation: Exponential and
Logarithmic Functions
Solve a Logarithmic Inequality
Solve log6 x > 3. Check your solution.
log6 x > 3
Original inequality
x > 63
Logarithmic to exponential
inequality
x > 216
Simplify.
Answer: The solution set is {x | x > 216}.
Solve a Logarithmic Inequality
Check
log6 x > 3
4 ?
log6 6 > 3
4 >3
Try 64 to see if it satisfies the inequality.
Original inequality
Substitute 64 for x.
log6 64 = 4
What is the solution to log3 x < 2?
A. {x | x < 9}
B. {x | 0 < x < 9}
C. {x | x > 9}
D. {x | x < 8}
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Solve Equations with Logarithms on
Each Side
Solve log4 x2 = log4 (4x – 3). Check your solution.
log4 x2 = log4 (4x – 3)
x2 = 4x – 3
x2 – 4x + 3 = 0
x =3
or
Property of Equality for
Logarithmic Functions
Subtract (4x – 3) from
each side.
(x – 3)(x – 1) = 0
x–3=0
Original equation
Factor.
x–1 =0
x =1
Zero Product Property
Solve each equation.
Solve Equations with Logarithms on
Each Side
Check
Substitute each value into the original
equation.
log4 x2 = log4 (4x – 3)
Original equation
2 ?
log4 3 = log4 (4(3) – 3) Substitute 3 for x.
log4 9 = log4 9 
log4 x2 = log4 (4x – 3)
Simplify.
Original equation
2 ?
log4 1 = log4 (4(1) – 3) Substitute 1 for x.
log4 1 = log4 1 
Simplify.
Answer: The solutions are 3 and 1.
What is the solution to log5 x2 = log5 (x + 6)?
A. –3 and 2
B. –1 and 6
C. 3 and –2
D. 1 and –6
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Solve Inequalities with Logarithms on
Each Side
Solve log7 (2x + 8) > log7 (x + 5). Check your
solution.
log7 (2x + 8) > log7 (x + 5)
2x + 8 > x + 5
x > –3
Original inequality
Property of Inequality for
Logarithmic Functions
Addition and Subtraction
Properties of Inequalities
We must exclude all values of x such that 2x + 8 ≤ 0 or
x + 5 ≤ 0. Thus the solution set is x > –4, x > –5, and
x > –3. This compound inequality simplifies to x > –3.
Answer: The solution set is x > –3.
What is the solution to log3 (2x + 6) > log3 (x + 2)?
A. x > –2
B. x > –4
C. x < –4
D. x < –2
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 9-2)
Main Ideas
Targeted TEKS
Key Concept: Product Property of Logarithms
Example 1: Use the Product Property
Key Concept: Quotient Property of Logarithms
Example 2: Use the Quotient Property
Example 3: Real-World Example
Key Concept: Writing Equations
Example 4: Power Property of Logarithms
Example 5: Solve Equations Using Properties of Logarithms
• Simplify and evaluate expressions using the
properties of logarithms.
• Solve logarithmic equations using the properties of
logarithms.
2A.11 The student formulates equations and
inequalities based on exponential and logarithmic
functions, uses a variety of methods to solve them, and
analyzes the solutions in terms of the situation.
(D) Determine solutions of exponential and
logarithmic equations using graphs, tables, and
algebraic methods.
Use the Product Property
Use log5 2 ≈ 0.4307 to approximate the value of
log5 250.
log5 2 = log5 (53 ● 2)
Replace 250 with 53 ● 2.
= log5 53 + log5 2
Product Property
= 3 + log5 2
Inverse Property of
Exponents and Logarithms
≈ 3 + 0.4307 or 3.4307 Replace log5 2 with
0.4307.
Answer: Thus, log5 250 is approximately 3.4307.
Given log2 3 ≈ 1.5850, what is the approximate value
of log2 96?
A. –3.415
B. 3.415
C. 5.5850
D. 6.5850
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Use the Quotient Property
Use log6 8 ≈ 1.1606 and log6 32 ≈ 1.9343 to
approximate the value of log6 4.
Replace 4 with
the quotient
.
= log6 32 – log6 8
Quotient Property
≈ 1.9343 – 1.1606 or 0.7737
log6 8 ≈ 1.1606 and
log6 32 ≈ 1.9343
Answer: Thus, log6 4 is approximately 0.7737.
Given log5 4 ≈ 0.8614 and log5 32 ≈ 2.1534, what is the
approximate value of log5 8?
A. –1.2920
B. 1.2920
C. 2.4999
D. 3.0148
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
SOUND The sound made by a lawnmower has a
relative intensity of 109 or 90 decibels. Would the
sound of ten lawnmowers running at the same
intensity be ten times as loud or 900 decibels?
Explain your reasoning.
Let L1 be the loudness of one lawnmower running.
L1 = 10 log10 109
Let L2 be the loudness of ten lawnmowers running.
L2 = 10 log10 (10 ● 109)
Then the increase in loudness is L2 – L1.
L2 – L1 = 10 log10 (10 ● 109) – 10 log10 109
Substitute for L1 and L2.
= 10 (log10 10 + log10 109) – 10 log10 109
Product Property
= 10 log10 10 + 10 log10 109 – 10 log10 109
Distributive Property
= 10 log10 10
Subtract.
= 10(1) or 10
Inverse Property of
Exponents and
Logarithms
Answer: No; the sound of ten lawnmowers is perceived
to be only 10 decibels louder than the sound of
one lawnmower, or 100 decibels.
SOUND The loudness L of a sound in decibels is given by L = 10
log10 R, where R is the sound’s relative intensity. The sound
made by fireworks has a relative intensity of 1014 or 140 decibels.
Would the sound of ten fireworks of that same intensity be ten
times as loud or 1400 decibels? Explain your reasoning.
A.
B.
C.
D.
No; the sound of ten fireworks is perceived
to be only 10 more decibels as loud as the
sound of one firework, or 150 decibels.
1.
No; the sound of ten fireworks is perceived
2.
to be 100 more decibels as loud as the
3.
sound of one firework, or 150 decibels.
Yes; the sound of ten fireworks is perceived4.
to be 10 times as loud.
No; the sound of ten fireworks is perceived
to be only 14 more decibels than one
firework.
A
B
C
D
0%
A
B
C
D
Power Property of Logarithms
Given that log5 6 ≈ 1.1133, approximate the value of
log5 216.
log5 216 = log5 63
Replace 216 with 63.
= 3 log5 6
Power Property
≈ 3(1.1133) or 3.3399
Replace log5 6 with
1.1133.
Answer: 3.3399
Given that log4 6 ≈ 1.2925, what is the approximate
value of log4 1296?
A. 0.3231
B. 2.7908
C. 5.1700
D. 6.4625
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Solve Equations Using Properties of
Logarithms
A. Solve 4 log2 x – log2 5 = log2 125.
Original equation
Power Property
Quotient Property
Property of Equality for
Logarithmic Functions
Multiply each side by 5.
x =5
Take the 4th root of
each side.
Solve Equations Using Properties of
Logarithms
Answer: 5
Solve Equations Using Properties of
Logarithms
B. Solve log8 x + log8 (x – 12) = 2.
log8 x + log8 (x – 12) = 2
Original equation
log8 x(x – 12) = 2
Product Property
x(x – 12) = 82
x2 – 12x – 64 = 0
(x + 4)(x – 16) = 0
Definition of logarithm
Subtract 64 from each side.
Factor.
Solve Equations Using Properties of
Logarithms
x + 4 = 0 or x – 16 = 0
x = –4
x = 16
Zero Product Property
Solve Equations Using Properties of
Logarithms
Check Substitute each value into the original equation.
?
log8 (–4) + log8 [(–4) – 12)] = 2
?
log8 (–4) + log8 (–16) = 2
Replace x with –4.
Product Property
Since log8 (–4) and log8(–16) are undefined, –4 is an
extraneous solution and must be eliminated.
?
log8 16 + log8 (16 – 12) = 2
?
log8 16 + log8 4 = 2
?
log8 (16 ● 4) = 2
Replace x with 16.
16 – 12 = 4
Product Property
Solve Equations Using Properties of
Logarithms
?
log8 64 = 2
2 =2
16 ● 4 = 64
Definition of logarithm
Answer: The only solution is x = 16.
A. What is the solution to the equation
2 log3 x – 2 log3 6 = log3 25?
A. 4
B. 18
C. 30
D. 144
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. What is the solution to the equation
log2 x + log2 (x – 6) = 4?
A. 11
B. 8
C. 8, –2
D. no solution
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 9-3)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Find Common Logarithms
Example 2: Real-World Example: Solve Logarithmic Equations
Example 3: Solve Exponential Equations Using Logarithms
Example 4: Solve Exponential Inequalities Using Logarithms
Key Concept: Change of Base Formula
Example 5: Change of Base Formula
• Solve exponential equations and inequalities using
common logarithms.
• Evaluate logarithmic expressions using the Change
of Base Formula.
• common logarithm
• Change of Base Formula
2A.11 The student formulates equations and
inequalities based on exponential and logarithmic
functions, uses a variety of methods to solve them, and
analyzes the solutions in terms of the situation.
(C) Determine the reasonable domain and range
values of exponential and logarithmic functions, as well
as interpret and determine the reasonableness of
solutions to exponential and logarithmic equations
and inequalities. (D) Determine solutions of
exponential and logarithmic equations using graphs,
tables, and algebraic methods.
Find Common Logarithms
A. Use a calculator to evaluate log 6 to four decimal
places.
Keystrokes: LOG
Answer: about 0.7782
6
ENTER
.7781512504
Find Common Logarithms
B. Use a calculator to evaluate log 0.35 to four
decimal places.
Keystrokes: LOG
.35
Answer: about –0.4559
ENTER
–.4559319556
A. Which value is approximately equivalent to log 5?
A. 0.3010
B. 0.6990
C. 5.0000
D. 100,000.0000
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. Which value is approximately equivalent to log
0.62?
A. –0.2076
B. 0.6200
C. 1.2076
0%
0%
A
B
D. 4.1687
A. A
B. 0% B
C. C
C
D. D
0%
D
Solve Logarithmic Equations
EARTHQUAKE The amount of energy E, in ergs,
that an earthquake releases is related to its Richter
scale magnitude M by the equation
log E = 11.8 + 1.5M. The San Fernando Valley
earthquake of 1994 measured 6.6 on the Richter
scale. How much energy did this earthquake
release?
log E = 11.8 + 1.5M
Write the formula.
log E = 11.8 + 1.5(6.6)
Replace M with 6.6.
log E = 21.7
Simplify.
10log E = 1021.7
Write each side using
10 as a base.
Solve Logarithmic Equations
E = 1021.7
Inverse Property of
Exponents and
Logarithms.
E ≈ 5.01 × 1021
Use a calculator.
Answer: The amount of energy released was about
5.01 × 1021 ergs.
EARTHQUAKE The amount of energy E, in ergs, that
an earthquake releases is related to its Richter scale
magnitude M by the equation log E = 11.8 + 1.5M. In
1999 an earthquake in Turkey measured 7.4 on the
Richter scale. How much energy did this earthquake
release?
0%
A. –7.29 ergs
B. –2.93 ergs
C. 22.9 ergs
D. 7.94 × 1022 ergs
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Solve Exponential Equations
Using Logarithms
Solve 5x = 62.
5x = 62
log 5x = log 62
x log 5 = log 62
Original equation
Property of Equality
for Logarithms
Power Property of
Logarithms
Divide each side by log 5.
x ≈ 2.5643
Answer: 2.5643
Use a calculator.
Solve Exponential Equations
Using Logarithms
Check
You can check this answer by using a
calculator or by using estimation. Since 52 = 25
and 53 = 125, the value of x is between 2 and 3.
Thus, 2.5643 is a reasonable solution. 
What is the solution to the equation 3x = 17?
A. 0.3878
0%
B. 2.5713
C. 2.5789
D. 5.6667
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Solve Exponential Inequalities
Using Logarithms
Solve 27x > 35x – 3.
27x > 35x – 3
Original
inequality
log 27x > log 35x – 3
Property of
Inequality for
Logarithmic
Functions
7x log 2 > (5x – 3) log 3
Power Property
of Logarithms
7x log 2 > 5x log 3 – 3 log 3 Distributive
Property
7x log 2 – 5x log 3 > – 3 log 3
Subtract 5x log 3
from each side.
Solve Exponential Inequalities
Using Logarithms
x(7 log 2 – 5 log 3) > –3 log 3
Distributive
Property
Divide each
side by 7 log 2
– 5 log 3.
Switch > to <
because
7 log 2 –
5 log 3 is
negative.
Use a
calculator.
Simplify.
Solve Exponential Inequalities
Using Logarithms
Check: Test x = 0.
27x > 35x – 3
7(0) ?
2
> 35(0) – 3
0 ?
2 > 3–3
Original inequality
Replace x with 0.
Simplify.

Negative Exponent
Property
Answer: The solution set is {x | x < 5.1415}.
What is the solution to 53x < 10x –2?
A. {x | x > –1.8233}
B. {x | x < 0.9538}
C. {x | x > –0.9538}
D. {x | x < –1.8233}
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Change of Base Formula
Express log3 18 in terms of common logarithms.
Then approximate its value to four decimal places.
Change of Base Formula
Use a calculator.
Answer: The value of log3 18 is approximately 2.6309.
What is log5 16 expressed in terms of common
logarithms and approximated to four decimal places?
A.
B.
C.
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 9-4)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Evaluate Natural Base Expressions
Example 2: Evaluate Natural Logarithmic Expressions
Example 3: Write Equivalent Expressions
Example 4: Solve Base e Equations
Example 5: Real-World Example: Solve Base e
Inequalities
Example 6: Solve Natural Log Equations and Inequalities
• Evaluate expressions involving the natural base
and natural logarithms.
• Solve exponential equations and inequalities using
natural logarithms.
• natural base, e
• natural base exponential function
• natural logarithm
• natural logarithmic function
2A.11 The student formulates equations and
inequalities based on exponential and logarithmic
functions, uses a variety of methods to solve them, and
analyzes the solutions in terms of the situation.
(D) Determine solutions of exponential and
logarithmic equations using graphs, tables, and
algebraic methods. (F) Analyze a situation modeled
by an exponential function, formulate an equation or
inequality, and solve the problem. Also addresses
TEKS 2A.11(A).
Evaluate Natural Base Expressions
A. Use a calculator to evaluate e0.5 to four decimal
places.
Keystrokes:
2nd
[ex] 0.5 ENTER
Answer: about 1.6487
1.648721271
Evaluate Natural Base Expressions
B. Use a calculator to evaluate e–8 to four decimal
places.
Keystrokes:
2nd
[ex] –8
Answer: about 0.0003
ENTER
.0003354626
A. What is e0.3 rounded to four decimal places.
A. 0.0498
B. 0.7408
C. 1.3499
D. 20.0855
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. What is e–2 rounded to four decimal places.
A. –7.3891
B. –0.1353
C. 0.1353
D. 7.3891
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Evaluate Natural Logarithmic Expressions
A. Use a calculator to evaluate ln 3 to four decimal
places.
Keystrokes:
LN
Answer: about 1.0986
3
ENTER
1.098612289
Evaluate Natural Logarithmic Expressions
B. Use a calculator to evaluate
places.
Keystrokes:
LN
1÷4
Answer: about –1.3863
to four decimal
ENTER
–1.386294361
A. What is ln 2 rounded to four decimal places?
A. –2.0000
B. –0.6931
0%
C. 0.6931
D. 2.0000
1.
2.
3.
4.
A
B
C
D
A
B
C
D
B. What is
rounded to four decimal places?
A. –2.000
B. –0.6931
0%
C. 0.6931
D. 0.5000
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Write Equivalent Expressions
A. Write an equivalent logarithmic equation for
ex = 23.
ex = 23 → loge 23 = x
Answer: ln 23 = x
Write Equivalent Expressions
B. Write an equivalent exponential equation for
ln x ≈ 1.2528.
ln x ≈ 1.2528 → loge x ≈ 1.2528
Answer: x ≈ e1.2528
A. What is ex = 6 written in an equivalent logarithmic
equation?
A. x = ln 6
0%
B. x = ln 6 – ln e
C. ln x = 6
D. ln e = x – 6
1.
2.
3.
4.
A
A
B
C
D
B
C
D
B. What is ln x = 2.25 written in an equivalent
exponential equation?
A. x2.25 = e
0%
B. x = e2.25
C. xe = 2.25
D. 2.25x = e
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Solve Base e Equations
Solve 3e–2x + 4 = 10.
3e–2x + 4 = 10
3e–2x = 6
Original equation
Subtract 4 from each side.
e–2x = 2
Divide each side by 3.
ln e–2x = ln 2
Property of Equality for
Logarithms
–2x = ln 2
Inverse Property of Exponents
and Logarithms
Divide each side by –2.
Solve Base e Equations
x ≈ –0.3466
Use a calculator.
Answer: The solution is about –0.3466.
Solve Base e Equations
Check
You can check this value by substituting
–0.3466 into the original equation or by finding
the intersection of the graphs of y = 3e–2x + 4
and y = 10.
What is the solution to the equation 2e–2x + 5 = 15?
A. –0.8047
B. –0.6931
C. 0.6931
D. 0.8047
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Solve Base e Inequalities
A. SAVINGS Suppose your deposit $700 into an
account paying 3% annual interest, compounded
continuously. What is the balance after 8 years?
A = Pert
Continuous compounding
formula
= 700e(0.03)(8)
Replace P with 700, r with 0.03
and t with 8.
= 700e0.24
Simplify.
≈ $889.87
Use a calculator.
Answer: The balance after 8 years would be $889.87.
Solve Base e Inequalities
B. How long will it take for the balance in your
account to reach at least $1200?
The balance is at least $1200.
A
≥
1200
Write an inequality.
Replace A with
700e(0.03)t.
Divide each side by 700.
Solve Base e Inequalities
Property of Inequality for
Logarithms
Inverse Property of
Exponents and
Logarithms
Divide each side by 0.03.
t ≥ 18.0
Use a calculator.
Answer: It will take at least 18.0 years for the balance
to reach $1200.
A. SAVINGS Suppose your deposit $700 into an
account paying 6% annual interest, compounded
continuously. What is the balance after 7 years?
A. $46,058.59
B. $46,680.43
0%
D
A
B
0%
C
D
C
D. $365.37
A
0%
A.
B.
0%
C.
D.
B
C. $1065.37
B. SAVINGS Suppose your deposit $700 into an
account paying 6% annual interest, compounded
continuously. How long will it take for the balance in
your account to reach at least $2500?
A. at least 1.27 years
B. at least 7.50 years
D. at least 124.93 years
0%
D
A
B
0%
C
D
C
A
0%
B
C. at least 21.22 years
A.
B.
0%
C.
D.
Solve Natural Log Equations and
Inequalities
A. Solve ln 3x = 0.5.
Original equation
Write each side using exponents and
base e.
Inverse Property of Exponents and
Logarithms
Divide each side by 3.
Use a calculator.
Answer: The solution is 0.5496. Check this solution
using substitution or graphing.
Solve Natural Log Equations and
Inequalities
B. Solve ln (2x – 3) < 2.5.
Original inequality
Write each side using
exponents and base e.
Inverse Property of Exponents
and Logarithms
Add 3 to each side.
Divide each side by 2.
Use a calculator.
Solve Natural Log Equations and
Inequalities
Answer: The solution is all numbers less than 7.5912
and greater than 1.5 since we must exclude all
values such that 2x – 3 ≤ 0. Check this
solution using substitution.
A. What is the solution to ln 2x = 0.7?
A. 0.0069
B. 1.0069
C. 2.0138
D. 2.5059
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. What is the solution to ln (x – 3) < 3?
A. x < 23.0855
B. x > 23.0855
C. 3 > x > 20.0855
D. 3 < x < 23.0855
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 9-5)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Exponential Decay of the Form
y = a(1 – r)t
Example 2: Exponential Decay of the Form y = ae–kt
Example 3: Standardized Test Example
Example 4: Exponential Growth of the Form y = aekt
• Using logarithms to solve problems involving
exponential decay.
• Use logarithms to solve problems involving
exponential growth.
• rate of decay
• rate of growth
2A.11 The student formulates equations and inequalities
based on exponential and logarithmic functions, uses a
variety of methods to solve them, and analyzes the
solutions in terms of the situation. (D) Determine
solutions of exponential and logarithmic equations
using graphs, tables, and algebraic methods.
(F) Analyze a situation modeled by an exponential
function, formulate an equation or inequality, and
solve the problem.
Exponential Decay of the Form y = a(1 – r)t
CAFFEINE A cup of coffee contains 130 milligrams
of caffeine. If caffeine is eliminated from the body at
a rate of 11% per hour, how long will it take for 90%
of the caffeine to be eliminated?
Explore
The problem gives the amount of caffeine
consumed and the rate at which the caffeine is
eliminated. It asks you to find the time it will
take for 90% of the caffeine to be eliminated
from a person’s body.
Plan
Use the formula y = a(1 – r)t. Let t be the
number of hours since drinking the coffee.
The amount remaining y is 10% of 130 or 13.
Exponential Decay of the Form y = a(1 – r)t
Solve y = a(1 – r)t
Exponential decay formula
Replace y with 13, a with 130,
and r with 0.11.
Divide each side by 130.
log 0.10 = log 0.89t
Property of Equality for
Logarithms
log 0.10 = t log 0.89
Power Property for Logarithms
Divide each side by log 0.89.
Use a calculator.
Exponential Decay of the Form y = a(1 – r)t
Answer: It will take approximately 20 hours for 90% of
the caffeine to be eliminated from a person’s
body.
Check
Use the formula to find how much of the
original 130 milligrams of caffeine would
remain after 20 hours.
y = a(1 – r)t
Exponential decay formula
y = 130(1 – 0.11)20
Replace a with 130, r with
0.11 and t with 20.
y ≈ 12.64
Ten percent of 130 is 13, so the answer seems
reasonable.
CAFFEINE A cup of coffee contains 130 milligrams
of caffeine. If caffeine is eliminated from the body at
a rate of 11% per hour, how long will it take for 80%
of this caffeine to be eliminated from a person’s
body?
A. 0.1 hours
B. 1.9 hours
0%
D
A
0%
A
B
0%
C
D
C
D. 15.2 hours
A.
B.
0%
C.
D.
B
C. 13.8 hours
Exponential Decay of the Form y = ae–kt
A. GEOLOGY The half-life of Sodium-22 is 2.6
years. What is the value of k and the equation of
decay for Sodium-22?
Exponential decay formula
Replace y with 0.5a and t with 2.6.
Divide each side by a.
In 0.5 = In e–2.6k
Property of Equality for Logarithmic
Functions
In 0.5 = –2.6k
Inverse Property of Exponents and
Logarithms
Divide each side by –2.6.
Exponential Decay of the Form y = ae–kt
0.2666 ≈ k
Use a calculator.
Answer: The value of k of Sodium-22 is 0.2666. Thus,
the equation for the decay of Sodium-22 is
y = ae–0.2666t, where t is given in years.
Exponential Decay of the Form y = ae–kt
B. A geologist examining a meteorite estimates that
it contains only about 10% as much Sodium-22 as it
would have contained when it reached the surface
of the Earth. How long ago did the meteorite reach
the surface of the Earth?
y = ae–0.2666t
0.1a = ae–0.2666t
Formula for the decay of
Sodium-22
Replace y with 0.1a.
0.1 = e–0.2666t
Divide each side by a.
ln 0.1 = ln e–0.2666t
Property of Equality for
Logarithms
Exponential Decay of the Form y = ae–kt
ln 0.1 = –0.2666t
Inverse Property for
Exponents and Logarithms
Divide each side by
–0.2666.
8.64 ≈ t
Use a calculator.
Answer: It reached the surface of Earth about 9 years
ago.
A. HEALTH The half-life of radioactive iodine used in
medical studies is 8 hours. What is the value of k for
radioactive iodine?
A. k = –0.0866
0%
B. k = –4.1589
C. k = 0.0866
D. k = 4.1589
1.
2.
3.
4.
A
B
C
D
A
B
C
D
B. HEALTH The half-life of radioactive iodine used in
medical studies is 8 hours. A doctor wants to know
when the amount of radioactive iodine in a patient’s
body is 20% of the original amount. When will this
occur?
A. about 0.05 hours later
B. about 0.39 hours later
C. about 2.58 hours later
D. about 18.58 hours later
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
TEST EXAMPLE The population of a city of one
million is increasing at a rate of 3% per year. If the
population continues to grow at this rate, in how
many years will the population have doubled?
A 4 years
C 20 years
B 5 years
D 23 years
Read the Test Item
You want to know when the population has doubled or is
2 million. Use the formula y = a(1 + r)t.
Solve the Test Item
y = a(1 + r)t
Exponential growth formula
2,000,000 = 1,000,000(1 + 0.03)t
Replace y with 2,000,000, a
with 1,000,000, and r with 0.03.
2 = (1.03)t
Divide each side by 1,000,000
and simplify.
ln 2 = ln(1.03)t
Property of Equality for
Logarithms
ln 2 = t ln 1.03
Power Property of Logarithms
Divide each side by ln 1.03.
t ≈ 23.45
Answer: D
Use a calculator.
The population of a city of 10,000 is increasing at a
rate of 5% per year. If the population continues to
grow at this rate, in how many years will the
population have doubled?
A. 10 years
B. 12 years
C. 14 years
D. 18 years
0%
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Exponential Growth of the Form y = aekt
POPULATION As of 2005, Nigeria had an estimated
population of 129 million people, and the United
States had an estimated population of 296 million
people. Assume that the populations of Nigeria and
the United States can be modeled by N(t) = 129e0.024t
and U(t) = 296e0.009t , respectively. According to these
models, when will Nigeria’s population be more than
the population of the United States?
You want to find t such that N(t) > U(t).
N(t) > U(t)
129e0.024t > 296e0.009t
Replace N(t) with
129e0.024t and U(t)
with 296e0.009t.
Exponential Growth of the Form y = aekt
ln 129e0.024t > ln 296e0.009t
ln 129 + ln e0.024t > ln 296 + ln e0.009t
ln 129 + 0.024t > ln 296 + 0.009t
Property of
Inequality for
Logarithms
Product
Property of
Logarithms
Inverse
Property of
Exponents and
logarithms
Exponential Growth of the Form y = aekt
ln 129 – ln 296 > –0.015t
Subtract ln 296
and 0.024t from
each side.
Divide each
side by –0.015.
55 < t
Answer: After 55 years or in
2060, Nigeria’s population
will be greater than the
population in the U.S.
Use a
calculator.
Interactive Lab: Exploring Exponential
and Logarithmic Functions
POPULATION As of 2000, Saudi Arabia had an estimated
population of 20.7 million people and the United States
had an estimated population of 278 million people. The
growth of the populations of Saudi Arabia and the United
States can be modeled by S(t) = 20.7e0.0327t and U(t) =
278e0.009t, respectively. According to these models, when
will Saudi Arabia’s population be more than the
population of the United States?
A. after 207 years or in the year 2207
0%
D
A
D. after 365 years or in the year 2365
A
B
0%
C
D
C
C. after 109 years or in the year 2109
A.
B.
0% C.
0%
D.
B
B. after 62 years or in the year 2062
Five-Minute Checks
Image Bank
Math Tools
Exponential and Logarithmic Functions
Exploring Exponential and Logarithmic
Functions
Lesson 9-1 (over Chapter 8)
Lesson 9-2 (over Lesson 9-1)
Lesson 9-3 (over Lesson 9-2)
Lesson 9-4 (over Lesson 9-3)
Lesson 9-5 (over Lesson 9-4)
Lesson 9-6 (over Lesson 9-5)
To use the images that are on the
following three slides in your own
presentation:
1. Exit this presentation.
2. Open a chapter presentation using a
full installation of Microsoft® PowerPoint®
in editing mode and scroll to the Image
Bank slides.
3. Select an image, copy it, and paste it
into your presentation.
(over Chapter 8)
A.
B.
C.
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Chapter 8)
A.
B.
C.
D. 21x
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Chapter 8)
Identify the type of function
represented by the graph.
A. absolute value
1.
2.
3.
4.
B. inverse variation
0%
C. quadratic
D. square root
A
B
C
D
A
B
C
D
(over Chapter 8)
A. s = –8
B. s = –12
C. s = 8
D. s = 12
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Chapter 8)
If y varies directly as x and y = 12 when x = –8, find
x when y = 4.
A. x = –24
0%
B. x = –6
C.
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Chapter 8)
If 2x – 3 = 32, then 2x =
A. 8.
0%
B. 29.
1.
2.
3.
4.
C. 128.
D. 256.
A
B
A
B
C
D
C
D
(over Lesson 9-1)
A. growth
A
B
0%
B
0%
A
B. decay
1.
2.
(over Lesson 9-1)
Determine whether the function y = 3(0.3)x
represents exponential growth or decay.
A. growth
B. decay
1.0% A
2. B
A
B
(over Lesson 9-1)
A. n20
0%
5
B. 4n
1.
2.
3.
4.
C.
A
B
C
D
D.
A
B
C
D
(over Lesson 9-1)
A. y = –2
B. y = –1
C.
D. y = 0
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-1)
The number of bees in a hive is
growing exponentially. Write an
exponential function to model
the population y of bees x days
after the sixth day.
A. y = 50(3.16)
0%
x
B. y = 15(3.162)x
C. y = 5(10)x
1.
2.
3.
4.
A
B
C
D
A
D. y = 0.05(3.16)x
B
C
D
(over Lesson 9-1)
If 16x + 1 = 322x, then 43x + 1 =
A. 16.
0%
B. 32.
1.
2.
3.
4.
C. 64.
D. 256.
A
B
A
B
C
D
C
D
(over Lesson 9-2)
A.
B.
C.
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-2)
Write log6 216 = 3 in exponential form.
A. 36 = 216
B. 63 = 216
C.
D.
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-2)
Evaluate 4log4 (x – 2).
A. 4
0%
B. 2
1.
2.
3.
4.
C. 4(x – 2)
A
B
C
D
D. x – 2
A
B
C
D
(over Lesson 9-2)
Solve log3 x < 3.
A. {x | 0 < x < 27}
B. {x | 0 < x < 9}
C. {x | 0 < x < 6}
D. {x | 0 < x < 1}
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-2)
Solve log4 (x2 – 30) = log4 x.
A. –6
B. –5
C. 6
D. 5
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-2)
The curve shown in the figure
represents a portion of the graph of
which of the following functions?
A. y = x2
0%
1.
2.
3.
4.
B. y = log2 x
C. y = logx 2
D. y = 2x
A
B
C
D
A
B
C
D
(over Lesson 9-3)
Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to
approximate the value of log3 32.
A. 10.0952
B. 7.5712
C. 3.1547
D. 2.3885
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-3)
Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to
approximate the value of log3
A. 1.4999
0%
B. 0.6666
C. –0.3155
D. –0.6309
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-3)
Solve log5 6 + 3log5 x = log5 48.
A. 2
0%
B. 3
1.
2.
3.
4.
C. 5
A
B
C
D
D. 8
A
B
C
D
(over Lesson 9-3)
Solve log2 (n + 4) + log2 n = 5.
A. 1
B. 4
C. 8
D. 9
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-3)
Solve log6 16 – 2log6 4 = log6 (x + 1) + log6
A. 31
B. 29
C. 3
D.
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-3)
Simplify 3log4 6 – log4 12 + 2log4 2 – log4 9.
A. log4 2
0%
B. log4 199
1.
2.
3.
4.
C. 5log4 1
D. log4 8
A
B
A
B
C
D
C
D
(over Lesson 9-4)
Use a calculator to evaluate log 3.4 to four decimal
places.
A. 0.5315
B. 0.6334
C. 1.5314
D. 1.6334
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-4)
Solve the equation 2x – 4 = 14. Round to four
decimal places.
A. 3.8073
0%
B. 4.8451
C. 6.2921
D. 7.8074
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-4)
Solve the inequality 42p – 1 > 11p. Round to four
decimal places.
A. {p | p > 1.0273}
0%
B. {p | p > 2.8712}
1.
2.
3.
4.
C. {p | p > 1.15146}
D. {p | p > 4.7296}
A
B
A
B
C
D
C
D
(over Lesson 9-4)
Express log4 (2.2)3 in terms of common logarithms.
Then approximate its value to four decimal places.
A.
B.
C.
D.
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-4)
Given x = 10log 7 and y = log7 7, which statement is
true?
A. x > y
0%
B. x < y
C. x = y
D. 7x = y
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-5)
Write an equivalent exponential equation for
ln 14.2 = x.
A. 14.2e = x
B. e14.2 = x
C. ex = 14.2
D. 14.2x = e
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-5)
Evaluate ln e2x – 3.
A. e
B. 2x – 3
C. e2x – 3
D. (2x – 3)e
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 9-5)
Solve the equation 6 + 4e–x = 12.
A. –0.4055
0%
B. –0.1761
1.
2.
3.
4.
C. 0.8385
A
B
C
D
D. 1.1925
A
B
C
D
(over Lesson 9-5)
Solve the inequality ln (2x + 3) > –3.
A. {x | x > 8.5428}
B. {x | x > 1.5249}
C. {x | x > –1.4751}
D. {x | x > –4.9539}
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 9-5)
You deposit $2000 in an account paying 4%
annual interest compounded continuously. Using
the continuous compounding formula A = Pert,
how long will it take for your money to double?
A. about 21.7 years
1.
2.
3.
4.
0%
B. about 17.3 years
C. about 5.8 years
D. about 3.5 years
A
B
C
D
A
B
C
D
(over Lesson 9-5)
A. –1.10
0%
1.
2.
3.
4.
B. –0.55
C. 0.20
D. 0.54
A
B
A
B
C
D
C
D
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