Algebra 2 This lesson needs new art
Lesson 8-1
Example 1 Graph Exponential Growth Functions
x
1
Sketch the graph of y = 3   . Then state the function’s domain and range.
 2
Make a table of values. Connect the points to sketch a smooth curve.
x
x
1
y = 3 
 2
–3
3 
–2
3 
–1
1
3 
 2
0
1
3  = 3
 2
1
3   = 1.5
2
1
3   = 0.75
 2
3
1
3   = 0.375
 2
1
 2
3
= 24
1
 2
2
= 12
1
=6
0
1
1
 2
2
3
The domain is all real numbers, and the range is all positive numbers.
Example 2 Graph Transformations
Graph each function. State the domain and range.
a.
y = 3x  3
The equation represents a translation of the graph of y = 3x three units down.
x
-3
-2
-1
0
1
2
3
y
y = 3x  3
3-3  3 = 2.963
3-2  3 = 2.889
3-1  3 = 2.667
30  3 = 2
31  3 = 0
32  3 = 6
33  3 = 24
x
O
Domain = {all real numbers}
Range = {y | y > 3}
b.
y=

1 x-2
2
2

The equation represents a transformation of the graph of y = 2x.
Graph y = 2x and transform the graph.
1
 a = 2: The graph is expanded.
 h = 2: The graph is translated
2 units right.
 k = 0: The graph is not translated vertically.
y
O
x
Domain = {all real numbers}
Range = {y | y > 0}
Real-World Example 3 Graph Exponential Growth Functions
In 1990, the population of Montana was 799,065. Since then, the population has grown by
approximately 1.2% annually. Draw a graph showing the population growth of Montana
from 1990 to 2000.
First, write an equation using a = 799,065, and r = 0.012.
y = 799,065(1.012) t
Then graph the equation.
[0, 600] scl: 25 by [0, 500,000,000}
scl: 20,000
Example 4 Graph Exponential Decay Functions
Graph each function. State the domain and range.
 1
a. y = 2  
 2
x
x
Graph black line.
1
y = 2 
 2
x
3
–3
1
2 
 2
–2
2 
–1
1
2 
 2
0
1
2  = 2
 2
1
2  = 1
2
1
2   = 0.5
 2
3
 1
2   = 0.25
 2
= 16
1
 2
2
=8
1
=4
0
1
1
 2
2
3
The domain is all real numbers, and the range is all positive real numbers.
1
b. y = 3  
4
x 2
+2
x
1
The equation represents a transformation of the graph of y =   .
4
Examine each parameter.
 a = 3: The graph is compressed.
 h = -2: The graph is translated
2 units left.
 k = 2: The graph is translated
2 units up.
The range is all real numbers, and the domain is all
real numbers greater than -3.
Algebra 2
Ignore green, use blue
and black
Lesson 8-2
Example 1 Solve Exponential Equations
Solve each equation.
a.
1
 
 2
n–1
1
 
2
n –1
= 16
= 16
(2–1)n – 1 = 24
2–n + 1 = 24
–n + 1 = 4
–n = 3
n = –3
Original equation
1
as 2–1 and 16 as 24 so each side as the same base.
2
Power of a Power
Property of Equality for Exponential Functions
Subtract 1 from each side.
Divide each side by –1.
Rewrite
b. 55n + 1 = 125n – 2
55n + 1 = 125n – 2
55n + 1 = (53)n - 2
55n + 1 = 53n – 6
5n + 1 = 3n – 6
2n + 1 = –6
2n = –7
n = –3.5
Original equation
Rewrite 125 as 53 so each side has the same base.
Power of a Power
Property of Equality for Exponential Functions
Subtract 3n from each side.
Subtract 1 from each side.
Divide each side by 2.
Real-World Example 2 Write an Exponential Function
SCIENCE Lee starts an experiment with 4500 bacteria cells. After 6 hours, there are 17,000
cells in his sample.
a. Write an exponential function that could be used to model the number of bacteria after
x hours if the number of bacteria changes at the same rate.
At the beginning of the experiment, time is 0 hours and there are 4500 bacteria cells. Thus, the yintercept, and the value of a, is 4500.
When x = 6, then number of bacteria cells is 17,000. Substitute these values into an exponential
function to determine the value of b.
6
y = abx
17,000 = 4500  b6
3.778  b6
Exponential function
Replace x with 6, y with 17,000, and a with 4500.
Divide each side by 4500.
3.778  b
1.248  b
Take the 6th root of each side.
Use a calculator.
An equation that models the number of bacteria is y = 4500(1.248)x.
b. How many bacteria cells can be expected in the sample after 10 hours?
y = 4500(1.248)x
= 4500(1.248)10
 41243.77
Modeling equation
Replace x with 10.
Use a calculator.
There will be approximately 41,244 bacteria cells after 10 hours.
Example 3 Compound Interest
An investment account pays 3.8% annual interest compounded monthly. If $5000 is invested
in this account, what will be the balance after 10 years?
Understand
Find the total amount in the account after 10 years.
Plan
Use the compound interest formula.
P = 5000, r = 0.038, n = 12, and t = 10
 r
A = P 1  
 n
Solve
nt
 0.038 
= 5000  1


12 
 7336.23
Check
Compound interest formula
12  10
P = 5000, r = 0.038, n = 12, t = 10
Use a calculator.
Graph the corresponding equation
y = 5000(1.0032)12t. Use CALC: value to find y
when x = 10.
The y-value 7336.228 is very close to 7336.23, so the answer is reasonable.
[0, 20] scl:1 by [4,000,
10,000] scl:1,000
Example 4 Solve Exponential Inequalities
Solve 9a – 5 < 729.
9a – 5 < 729
(3 )
< 36
32a – 10 < 36
2a – 10 < 6
2a < 16
a <8
2 a–5
Original inequality
Rewrite 9 as 3 2 and 729 as 3 6 so each side has the same base.
Power of a Power
Property of Inequality for Exponential Functions
Add 10 to each side.


Divide each side by 2.
Algebra 2
Lesson 8-4
Example 1 Solve a Logarithmic Equation
Solve log 1 a =
8
log 1 a =
8
2
3
2
3
.
Original equation
2
13
a =  
8
Definition of logarithm
2
1 3 3
a =   

2  

2
1 
1
a =   or
2 
4
1
8
1 3
2 
=  
Power of a Power
Standardized Test Example 2
Solve log6 (a2 – 15) = log6 (2a). Check your solution.
log6 (a2 – 15) = log6 (2a)
a2 – 15 = 2a
2
a – 2a – 15 = 0
(a – 5)(a + 3) = 0
a – 5 = 0 or
a+3=0
a = 5 or
a = –3
CHECK
Original equation
Property of Equality for Logarithmic Functions
Subtract 2a from each side.
Factor.
Zero Product Property
Solve each equation.
Substitute each value into the original equation.
log6 (52 – 15) = log6 [2(5)]
log6 10 = log6 10 
Substitute 5 for a.
Simplify.
log6 [(–3)2 – 15] = log6 [2(–3)]
Substitute –3 for a.
Since log6 (–6) is undefined, –3 is an extraneous solution and must be eliminated. Thus, the
solution is 5.
Example 3 Solve a Logarithmic Inequality
Solve log7 x > 3. Check your solution.
log7 x > 3
x > 73
x > 343
Original inequality
Logarithmic to exponential inequality.
Simplify.
The solution set is {x | x > 343}.
CHECK
Try 2401 to see if it satisfies the inequality.
log7 x > 3
Original inequality.
?
log7 (2401)  3
4>3 
Substitute 2401 for x.
log7 2401 = 4 because 74 = 2401.
Example 4 Solve Inequalities with Logarithms on Each Side
Solve log3 (5x + 7) < log3 (3x + 13).
log3 (5x + 7) < log3 (3x + 13)
5x + 7 < 3x + 13
2x < 6
x<3
Original inequality
Property of Inequality for Logarithmic Functions
Subtract Property of Inequalities
Divide each side by 2.
We must exclude all values of x such that 5x + 7  0 or 3x + 13  0. Thus, the solution set is
7
13
7
x > - , x > - , and x < 3. This compound inequality simplifies to - < x < 3. The solution set is
5
3
5
7
{x| -5 < x < 3}.
Algebra 2
Lesson 8-4
Example 1 Solve a Logarithmic Equation
Solve log 1 a =
8
log 1 a =
8
2
3
2
.
Original equation
3
2
13
a =  
8
Definition of logarithm
2
1 3 3
a =   

2  

1
8
1 3
2 
=  
1 2
1
a =   or
2 
4
Power of a Power
Standardized Test Example 2
Solve log6 (a2 – 15) = log6 (2a). Check your solution.
log6 (a2 – 15) = log6 (2a)
a2 – 15 = 2a
2
a – 2a – 15 = 0
(a – 5)(a + 3) = 0
a – 5 = 0 or
a+3=0
a = 5 or
a = –3
CHECK
Original equation
Property of Equality for Logarithmic Functions
Subtract 2a from each side.
Factor.
Zero Product Property
Solve each equation.
Substitute each value into the original equation.
log6 (52 – 15) = log6 [2(5)]
log6 10 = log6 10 
Substitute 5 for a.
Simplify.
log6 [(–3)2 – 15] = log6 [2(–3)]
Substitute –3 for a.
Since log6 (–6) is undefined, –3 is an extraneous solution and must be eliminated. Thus, the
solution is 5.
Example 3 Solve a Logarithmic Inequality
Solve log7 x > 3. Check your solution.
log7 x > 3
x > 73
x > 343
Original inequality
Logarithmic to exponential inequality.
Simplify.
The solution set is {x | x > 343}.
CHECK
Try 2401 to see if it satisfies the inequality.
log7 x > 3
Original inequality.
?
log7 (2401)  3
4>3 
Substitute 2401 for x.
log7 2401 = 4 because 74 = 2401.
Example 4 Solve Inequalities with Logarithms on Each Side
Solve log3 (5x + 7) < log3 (3x + 13).
log3 (5x + 7) < log3 (3x + 13)
5x + 7 < 3x + 13
2x < 6
x<3
Original inequality
Property of Inequality for Logarithmic Functions
Subtract Property of Inequalities
Divide each side by 2.
We must exclude all values of x such that 5x + 7  0 or 3x + 13  0. Thus, the solution set is
7
13
7
x > - , x > - , and x < 3. This compound inequality simplifies to - < x < 3. The solution set is
5
3
5
7
{x| -5 < x < 3}.
log3 (5x + 7) < log3 (3x + 13)
5x + 7 < 3x + 13
2x < 6
x<3
Original inequality
Property of Inequality for Logarithmic Functions
Subtract Property of Inequalities
Divide each side by 2.
We must exclude all values of x such that 5x + 7  0 or 3x + 13  0. Thus, the solution set is
7
13
7
x > -5, x > - 3 , and x < 3. This compound inequality simplifies to -5 < x < 3. The solution set is
7
{x| -5 < x < 3}.
Algebra 2
Lesson 8-5
Example 1 Use the Product Property
Use log3 2 ≈ 0.6310 to approximate the value of log3 54.
log3 54 = log3 (33  2)
= log3 33 + log3 2
= 3 + log3 2
= 3 + 0.6310 or 3.6310
Replace 54 with 27  2 or 33  2.
Product Property
Inverse Property of Exponents and Logarithms
Replace log3 2 with 0.6310.
Thus, log3 54 is approximately 3.6310.
Real-World Example 2 Quotient Property
Use log5 2 ≈ 0.4307 and log5 7 ≈ 1.2091 to approximate log5 3.5.
log5 3.5 = log5
7
2
Replace 3.5 with the quotient
7
.
2
= log5 7 – log5 2
Quotient Property
≈ 1.2091 – 0.4307 or 0.7784 log5 7 = 1.2091 and log5 2 = 0.4307
Thus, log5 3.5 is approximately 0.7784.
CHECK Using the definition of logarithm and a calculator, 50.7784 ≈ 3.5.
Example 3 Power Property of Logarithms
Given log3 7 ≈ 1.7712, approximate the value of log3 49.
log3 49 = log3 (72)
= 2 log3 7
≈ 2(1.7712) or 3.5424
Replace 49 with 72.
Power Property
Replace log3 7 with 1.7712.
Thus, log3 49 is approximately 3.5424.
Example 4 Solve Equations Using Properties of Logarithms
Solve 2 log2 x + log2 3 = log2 27.
2 log2 x + log2 3 = log2 27
log2 x2 + log2 3 = log2 27
log2 (x2  3) = log2 27
3x2 = 27
x2 = 9
x=±3
CHECK
Original equation
Power Property
Product Property
Property of Equality for Logarithmic Functions
Divide each side by 3.
Take the square root of each side.
Substitute each value into the original equation.
2 log2 x + log2 3 = log2 27
?
2 log2 (3) + log2 3  log2 27
?
log2 (32) + log2 3  log2 27
?
log2 (3  3)  log2 27
log2 27 = log2 27
2
2 log2 x + log2 3 = log2 27
?
2 log2 (–3) + log2 3  log2 27
Since log2 (–3) is undefined, –3
is an extraneous solution and must
be eliminated.
The only solution is 3.
Algebra 2
Lesson 8-6
Example 1 Find Common Logarithms
Use a calculator to evaluate each expression to four decimal places.
a. log 19
KEYSTROKES: LOG 19 ENTER 1.27875360095
b. log 0.75
KEYSTROKES: LOG 0.75 ENTER -.124938736608 about –0.1249
Example 3 Solve Exponential Equations Using Logarithms
Solve 2a+3 = 34 . Round to the nearest ten-thousandth.
2a + 3 = 34
log 2a + 3 = log 34
(a + 3) log 2 = log 34
a+3=
a=
Original equation
Property of Equality for Logarithmic Functions
Power Property of Logarithms
log 34
log 2
log 34
log 2
Divide each side by log 2.
–3
Subtract 3 from each side.
about 1.2788
a≈
1.5315
0.3010
–3
Use a calculator.
a ≈ 2.0880
The solution is approximately 2.0880.
CHECK
You can check this answer using a calculator or by using estimation. Since 25 = 32 and 32 is
close to 34, the value of x is slightly greater than 2 since 2 + 3 = 5. Thus, 2.0880 is a
reasonable solution.
Example 4 Solve Exponential Inequalities Using Logarithms
Solve 5b – 1 ≥ 106 – b.
5b – 1 ≥ 106 – b
log 5b – 1 ≥ log 106 – b
(b – 1) log 5 ≥ (6 – b) log 10
b log 5 – log 5 ≥ 6 log 10 – b log 10
b log 5 + b log 10 ≥ log 5 + 6 log 10
b(log 5 + log 10) ≥ log 5 + 6 log 10
b≥
b≥
log 5  6log 10
log 5  log10
0.6990  6 (1)
0.6990  1
b ≥ 3.9429
Original inequality
Property of Inequality for Logarithmic Functions
Power Property of Logarithms
Distributive Property
Add b log 10 and log 5 to each side.
Distributive Property
Divide each side by log 5 + log 10.
Use a calculator.
Use a calculator.
The solution set is {b | b ≥ 3.9429}.
CHECK
Test b = 4.
5b – 1 ≥ 106 – b
Original inequality
?
54 – 1  106 – 4
?
5  10
125 ≥ 100 
3
2
Replace b with 4.
Simplify.
Simplify.
Example 5 Change of Base Formula
Express log7 58 in terms of common logarithms. Then approximate its value to four decimal places.
log7 58 =
log10 58
Change of Base Formula
log10 7
≈ 2.0867
Use a calculator.
The value of log7 58 is approximately 2.0867.
Algebra 2
Lesson 8-7
Example 1 Write Equivalent Expressions
Write each exponential equation in logarithmic form.
a. e–x = 2
b x = e0.35
–x
e = 2  loge 2 = -x
x = e0.35  loge x = 0.35
ln 2 = -x
ln x = 0.35
Example 2 Write Equivalent Expressions
Write each logarithmic equation in exponential form.
a.
ln x = 0.6742
ln x = 0.6742 
b.
loge x = 0.6742
x = e0.6742
ln 22 = x
ln 22 = x 
loge 22 = x
22 = ex
Example 3 Simplify Expressions with e and the Natural Log
Write each expression as a single logarithm.
a. 4 ln 9 – ln 27
4 ln 9 – ln 27 = ln 94 – ln 27
Power Property of Logarithms
4
= ln
9
27
Quotient Property of Logarithms
= ln 243
= ln 35
= 5 ln 3
Simplify.
35 = 243
Power Property of Logarithms
1
b. ln 80 + 2 ln + ln y
4
1
1
ln 80 + 2 ln + ln y = ln 80 + ln + ln y
4
16
= ln 80 
= ln 5x
1
16
y
Power Property of Logarithms
Product Property of Logarithms
Simplify.
Example 4 Solve Base e Equations
Solve –2e5x + 10 = 6.
–2e5x + 10 = 6
–2e5x = –4
e5x = 2
ln e5x = ln 2
5x = ln 2
x=
1
5
Original equation
Subtract 10 from each side.
Divide each side by –2.
Property of Equality for Logarithms
Inverse Property of Exponents and Logarithms
1
Multiply each side by .
5
Use a calculator.
(ln 2)
x ≈ 0.1386
The solution is about 0.1386.
Example 5 Solve Natural Log Equations and Inequalities
a. ln (6x – 3) + 3 = 10
ln (6x – 3) + 3 = 10
ln (6x – 3) = 7
eln (6x – 3) = e7
6x – 3 = e7
6x = e7 + 3
Original equation
Subtract 3 from each side.
Write each side using exponents and base e.
Inverse Property of Exponents and Logarithms
Add 3 to each side.
e7  3
x=
6
x ≈ 183.2722
Divide each side by 6.
Use a calculator.
The solution is about 183.2722. Check this solution using substitution or by graphing.
b. ln (3x + 2) < 5
ln (3x + 2) < 5
eln (3x + 2) < e5
3x + 2 < e5
3x < e5 – 2
x<
Original inequality
Write each side using exponents and base e.
Inverse Property of Exponents and Logarithms
Subtract 2 from each side.
e5 – 2
3
x < 48.8044
Divide each side by 3.
Use a calculator.
Recall that (3x + 2) must be greater than 0 in order for ln (3x + 2) to be defined. Therefore, you
must solve 3x + 2 > 0.
3x + 2 > 0
3x > –2
x>–
Inequality to solve
Subtract 2 from each side.
2
Divide each side by 3.
3
The solution is –
2
3
< x < 48.8044.
Real-World Example 6 Solve Base e Inequalities
BANKING Suppose you deposit $5000 in an account paying 4% annual interest,
compounded continuously.
a. What is the balance after 8 years?
A


=
=
=

rt
Continuous compounding formula
Replace P with 5000, r with 0.04, and t with 8.
Simplify.
Use a calculator.
Pe
5000e(0.04 · 8)
0.32
5000e
6885.64

The balance
after 8 years would be $6885.64.
If the account was earning simple interest, the formula for the interest would be I
= Prt. In that case, the interest would be I = 5000(0.04)(8) or $1600.
Continuously compounded interest should be greater than simple interest at the
same rate. Thus the solution $6885.64 is reasonable.
CHECK
b. How long will it take for the balance in your account to reach at least $10,000?
Words
The balance is at least $10,000.
Variable(s)
Let A represent the amount in the account.
Inequality
A  10,000

 10,000
2
 ln 2
 ln 2
ln 2
t 
0.04
t  17.33
5000e 0.04t
e 0.04t
ln e 0.04t
0.04t
 
 
 

0.04t
Let A = 5000e
.
Divide both sides by 5000.
Property of Equality for Logarithms.
Inverse Property of Exponents and Logarithms

Divide each side by 0.04.
Use a calculator.
It will take atleast 17.33 years for the balance to reach $10,000.


Algebra 2
Lesson 8-8
Real-World Example 1 Exponential Decay
SCIENCE A particular chemical must be added to a swimming pool at regular intervals
because it is released from the water into the air at the rate of 5% per hour. Sixteen ounces
of the chemical is added at 8 am. At what time will
three–fourths of this chemical be gone from the pool?
Use the formula y = a(1 – r)t. Let t be the number of hours since adding the chemical. The initial
amount a is 16 ounces, and the percent of decrease r is 0.05. The amount remaining y is one–
fourth of 16 or 4.
y = a(1 – r)t
4 = 16(1 – 0.05)t
0.25 = (0.95)t
log 0.25 = log (0.95)t
log 0.25 = t log (0.95)
log 0.25
log 0.95
=t
27.0268 ≈ t
Exponential decay formula
Replace y with 4, a with 16, and r with 0.05.
Divide each side by 16.
Property of Equality for Logarithms
Product Property for Logarithms
Divide each side by log 0.95.
Use a calculator.
Since 27.0268 ≈ 27, it will take approximately 27 hours for three–fourths of the chemical to be
eliminated. The time will be 11 am the next day.
Real-World Example 2 Carbon Dating
SCIENCE The half–life of a radioactive substance is the time it takes for half of the atoms of
the substance to decay. Each element has a unique half–life. Radon–222 has a half–life of
about 3.8 days, while thorium–234 has a half–life of about 24 days. Find the value of k for
each element and compare their equations for decay.
The equations will be of the form y = ae–kt, where t is in days. To determine the constant k for
each element, let a be the initial amount of the substance. The amount y that remains after t days
of the half–life is then represented by 0.5a. Use this idea to find the value of k for each element
and then to write their equations.
Radon–222
y = ae–kt
0.5a = ae–k(3.8)
0.5 = e–3.8k
ln 0.5 = ln e–3.8k
ln 0.5 = –3.8k
ln 0.5
–3.8
=k
Exponential decay formula
Replace y with 0.5a and t with 3.8.
Divide each side by a.
Property of Equality for Logarithmic Functions
Inverse Property of Exponents and Logarithms
Divide each side by –3.8.
0.1824 ≈ k
Use a calculator.
Thorium–234
y = ae–kt
0.5a = ae–k(24)
0.5 = e–24k
ln 0.5 = ln e–24k
ln 0.5 = –24k
Exponential decay formula
Replace y with 0.5a and t with 24.
Divide each side by a.
Property of Equality for Logarithmic Functions
Inverse Property of Exponents and Logarithms
ln 0.5
–24
=k
0.0289 ≈ k
Divide each side by –24.
Use a calculator.
The equations for radon–222 and thorium–234 are y = ae–0.1824t and y = ae–0.0289t, respectively.
For both equations, t represents time in days. In comparing the equations, it appears that the
longer the half–life, the smaller the value of k.
Real-World Example 3 Continuous Exponential Growth
INVESTMENTS Sue invests $1000 at 5% interest compounded continuously and Norma
invests $1250 at 3.5% interest compounded continuously. When interest is compounded
continuously, the amount A in an account after t years is found using the formula A = Pert,
where P is the amount of principal and r is the annual interest rate. In how many years will
Sue’s account be greater than Norma’s account?
Since Sue’s interest rate is greater than Norma’s, it seems likely that Sue’s account will
eventually be greater than Norma’s. You need to write a function for Sue’s account and for
Norma’s account using the formula and then write an inequality. Let S be the amount in Sue’s
account and N be the amount in Norma’s account.
A = Pert
S = 1000e0.05t
Sue’s principal is 1000 and
rate is 5% or 0.05.
A = Pert
N = 1250e0.035t
Norma’s principal is 1250 and
rate is 3.5% or 0.035.
You want to find t such that S > N.
1000e0.05t > 1250e0.035t
ln 1000e0.05t > ln 1250e0.035t
ln 1000 + ln e0.05t > ln 1250 + ln e0.035t
ln 1000 + 0.05t > ln 1250 + 0.035t
ln 1000 + 0.015t > ln 1250
0.015t > ln 1250 – ln 1000
t >
t ≥ 14.88
ln 1250 – ln 1000
0.015
S>N
Property of Inequality for Logarithms
Product Property of Logarithms
Inverse Property of Exponents and Logarithms
Subtract 0.035t from each side.
Subtract ln 1000 from each side.
Divide each side by 0.015.
Use a calculator