Exponential FunctionsDefine an exponential function.Graph exponential functions.Use transformations on exponential functions.Define simple interest.Develop a compound interest formula. Understand the number e.
SECTION 4.1
1
2
3
4
5
6
EXPONENTIAL FUNCTION
A function f of the form
is called an exponential function with base a. Its domain is (–∞, ∞).
f x ax , a 0 and a 1,
EXAMPLE 1 Evaluating Exponential Functions
a. Let f x 3x 2. Find f 4 .
b. Let g x 210x. Find g 2 .
c. Let h x 1
9
x
. Find h 3
2
.
d. Let F(x) = 4x. Find F(3.2).
EXAMPLE 1 Evaluating Exponential Functions
Solution
4 2 2a. 94 3 3f
22
1 1b. 2 10 2 2 0.02
10 1002g
33
1 2
3
22
1c. 9 9
3
227
9h
d. F(3.2) = 43.2 ≈ 84.44850629
RULES OF EXPONENTS
Let a, b, x, and y be real numbers with a > 0 and b > 0. Then
,x y x ya a a
,x
x yy
aa
a
,x x xab a b
,yx xya a
0 1,a
1 1.
xx
xa
a a
EXAMPLE 2Graphing an Exponential Function with Base a > 1 – Exponential Growth
Graph the exponential function
Solution
Make a table of values.
f x 3x.
Plot the points and draw a smooth curve.
EXAMPLE 2Graphing an Exponential Function with
Base a > 1
Solution continued
This graph is typical for exponential functions when a > 1.
EXAMPLE 3Graphing an Exponential Function with Base 0 < a < 1 – Exponential Decay
Sketch the graph of
Solution
Make a table of values.
1.
2
x
y
Plot the points and draw a smooth curve.
EXAMPLE 3Graphing an Exponential Function with
Base 0 < a < 1
Solution continued
As x increases in the positive direction, y decreases towards 0.
PROPERTIES OF EXPONENTIAL FUNCTIONS
Let f (x) = ax, a > 0, a ≠ 1.
1. The domain of f (x) = ax is (–∞, ∞).
2. The range of f (x) = ax is (0, ∞); the entire graph lies above the x-axis.
3. For a > 1, Exponential Growth (i) f is an increasing function, so the graph rises to
the right.
(ii) as x → ∞, y → ∞.
(iii) as x → –∞, y → 0.
4. For 0 < a < 1, - Exponential Decay (i) f is a decreasing function, so the graph falls to the
right.(ii) as x → – ∞, y → ∞.
(iii) as x → ∞, y → 0.
5. The graph of f (x) = ax has no x-intercepts, so it never crosses the x-axis. No value of x will cause f (x) = ax to equal 0.
6. The graph of is a smooth and continuous curve, and it passes through the points
7. The x-axis is a horizontal asymptote for every exponential function of the form f (x) = ax.
TRANSFORMATIONS ON EXPONENTIAL
FUNCTION f (x) = ax Transformation Equation Effect on Equation
HorizontalShift
y = ax+b
Shift the graph of y = ax, |b| units(i) left if b > 0.(ii) right if b < 0.
VerticalShift
y = ax + b
Shift the graph of y = ax, |b| units(i) up if b > 0.(ii) down if b < 0.
TRANSFORMATIONS ON EXPONENTIAL
FUNCTION f (x) = ax Transformation Equation Effect on Equation
Stretching or Compressing(Vertically)
y = cax
Multiply the y coordinates by c. The graph of y = ax is vertically(i) stretched if c > 1.(ii) compressed if 0 < c < 1.
TRANSFORMATIONS ON EXPONENTIAL
FUNCTION f (x) = ax Transformation Equation Effect on Equation
Reflection y = –ax Reflect the graph of y = ax in the x-axis.
Reflect the graph of y = ax in the y-axis.
y = a–x
EXAMPLE 6 Sketching Graphs
Use transformations to sketch the graph of each function.
3 4xf x a.
State the domain and range of each function and the horizontal asymptote of its graph.
1 3xf x b.
3xf x c. 3 2xf x d.
EXAMPLE 6 Sketching Graphs
Solution
Domain: (–∞, ∞)
Range: (–4, ∞)
Horizontal Asymptote: y = –4
3 4xf x a.
EXAMPLE 6 Sketching Graphs
Solution continued
Domain: (–∞, ∞)
Range: (0, ∞)
Horizontal Asymptote: y = 0
1 3xf x b.
EXAMPLE 6 Sketching Graphs
Solution continued
Domain: (–∞, ∞)
Range: (–∞, 0)
Horizontal Asymptote: y = 0
3xf x c.
EXAMPLE 6 Sketching Graphs
Solution continued
Domain: (–∞, ∞)
Range: (–∞, 2)
Horizontal Asymptote: y = 2
3 2xf x d.
General Exponential Growth/Decay Model
Original amount
Rate of decay (r < 0),Growth (r > 0)
Number of time periods
Amount after t time periods
Compound interest is the interest paid on both the principal and the accrued (previously earned) interest. It is an application of exponential growth.
Interest that is compounded annually is paid once a year. For interest compounded annually, the amount A in the account after t years is given by
COMPOUND INTEREST – Growth
Amount after t time periods
Original amount
Rate of decay (r < 0),Growth (r > 0)
Number of time periods
EXAMPLE 2 Calculating Compound Interest
Juanita deposits $8000 in a bank at the interest rate of 6% compounded annually for five years.
a. How much money will she have in her account after five years?
b. How much interest will she receive?
EXAMPLE 2 Calculating Compound Interest
Solution
a. Here P = $8000, r = 0.06, and t = 5.
b. Interest = A P = $10,705.80 $8000 = $2705.80.
COMPOUND INTEREST FORMULA
A = amount after t yearsP = principalr = annual interest rate (expressed as a decimal)n = number of times interest is compounded
each yeart = number of years
1nt
rA P n
EXAMPLE 3Using Different Compounding Periods to Compare Future Values
If $100 is deposited in a bank that pays 5% annual interest, find the future value A after one year if the interest is compounded
(i) annually.(ii) semiannually.(iii) quarterly.(iv) monthly.(v) daily.
EXAMPLE 3Using Different Compounding Periods to Compare Future Values
(i) Annual Compounding:
1
1
1 $0.05 1 00 05 00 .
ntr
A Pn
A
Solution
In the following computations, P = 100, r = 0.05 and t = 1. Only n, the number of times interest is compounded each year, changes. Since t = 1, nt = n(1) = n.
EXAMPLE 3Using Different Compounding Periods to Compare Future Values
(iii) Quarterly Compounding:4
4
10
14
10
$105.0.05
40 9
rA P
A
(ii) Semiannual Compounding:2
2
10
1
1 $105.060
2
.050
rA P
n
A
EXAMPLE 3Using Different Compounding Periods to Compare Future Values
(iv) Monthly Compounding:1
12
2
1012
112
1 $10505
..0
120
rA P
A
(v) Daily Compounding:3
365
65
100.
1365
1 $1365
. 305
05 10
rA P
A
EXAMPLE 8 Bacterial Growth
A technician to the French microbiologist Louis Pasteur noticed that a certain culture of bacteria in milk doubles every hour. If the bacteria count B(t) is modeled by the equation
B t 20002t ,
a. the initial number of bacteria,b. the number of bacteria after 10 hours; andc. the time when the number of bacteria will be
32,000.
with t in hours, find
EXAMPLE 8 Bacterial Growth
00 2000 2 200 00 1 2000B B
a. Initial size
10b. 2000 2 2,10 048,000B
32,000 2000 2
16 2
t
t
c. Find t when B(t) = 32,000
24 2t
4 t
After 4 hours, the number of bacteria will be 32,000.
Solution
THE VALUE OF e
The value of e to 15 places is e = 2.718281828459045.
gets closer and closer to a fixed number. This irrational number is denoted by e and is sometimes called the Euler number.
As h gets larger and larger,
11
h
h
CONTINUOUS COMPOUND FORMULA
A = amount after t yearsP = principalr = annual rate (expressed as a decimal)t = number of years
ertA P
EXAMPLE 4 Calculating Continuous Compound Interest
Find the amount when a principal of $8300 is invested at a 7.5% annual rate of interest compounded continuously for eight years and three months.Solution
P = $8300 and r = 0.075. Convert eight years and three months to 8.25 years.
0.07 85 .25
$15,409.
$8300
83
rtA Pe
A e
EXAMPLE 5 Calculating the Amount of Repaying a Loan
How much money did the government owe DeHaven’s descendants for 213 years on a $450,000 loan at the interest rate of 6%?
Solution
a. With simple interest,
0.0$450,00
1
0 1
$6.201 million.
6 213
A P Prt P rt
A
EXAMPLE 5 Calculating the Amount of Repaying a Loan
Solution continued
b. With interest compounded yearly, 1
11
2 31 1
$1.105 10
$450,00
$110.500 millio
0.
n.
0 06t
A P r
A
c. With interest compounded quarterly, 4 4 21
11
30.0
1 14 4
$1.45305 10
$145.30
$450,
5 billi
0
o .
600
n
tr
A P
A
EXAMPLE 5 Calculating the Amount of Repaying a Loan
Solution continued
d. With interest compounded continuously,
Notice the dramatic difference between quarterly and continuous compounding and the dramatic difference between simple interest and compound interest.
2130.0
11
6
$1.5977 1
$450,00
0
$159.77 billion.
0rtA Pe e
A
THE NATURAL EXPONENTIAL FUNCTION
with base e is so prevalent in the sciences that it is often referred to as the exponential function or the natural exponential function.
xf x e
The exponential function
EXAMPLE 6 Sketching a Graph
Use transformations to sketch the graph of
Solution
Start with the graph of y = ex.
EXAMPLE 6 Sketching a Graph
Use transformations to sketch the graph of
Solution coninued
Shift the graph of y = ex one unit right.
EXAMPLE 6 Sketching a Graph
Use transformations to sketch the graph of
Solution continued
Shift the graph of y = ex – 1 two units up.
MODEL FOR EXPONENTIALGROWTH OR DECAY
0ktA t A e
A(t) = amount at time t A0 = A(0), the initial amount k = relative rate of growth (k > 0) or decay
(k < 0) t = time
EXAMPLE 7 Modeling Exponential Growth and Decay
In the year 2000, the human population of the world was approximately 6 billion and the annual rate of growth was about 2.1%. Using the model on the previous slide, estimate the population of the world in the following years.
a. 2030b. 1990
EXAMPLE 7
300.0213
11.2 5 3
6
66
0
6
A e
a. The year 2000 corresponds to t = 0. So A0 = 6 (billion), k = 0.021, and 2030 corresponds to t = 30.
Solution
The model predicts that if the rate of growth is 2.1% per year, over 11.26 billion people will be in the world in 2030.
Modeling Exponential Growth and Decay
EXAMPLE 7
0.021 10
4.863505
6
5
10A e
b. The year 1990 corresponds to t = 10.
Solution
The model predicts that the world had over 4.86 billion people in 1990. (The actual population in 1990 was 5.28 billion.)
Modeling Exponential Growth and Decay
Logarithmic Functions
Define logarithmic functions.Inverse FunctionsEvaluate logarithms.Rules of LogarithmsFind the domains of logarithmic functions.Graph logarithmic functions.Use logarithms to evaluate exponential equations.
SECTION 4.3
1
2
3
4
5
6
7
DEFINITION OF THELOGARITHMIC FUNCTION
For x > 0, a > 0, and a ≠ 1,
y loga x if and only if x ay .
The function f (x) = loga x, is called the logarithmic function with base a.
The logarithmic function is the inverse function of the exponential function.
Inverse Functions
Certain pairs of one-to-one functions “undo” one another. For example, if
5( ) 8 5 and ( ) ,
8x
x x x f g
then
855)10(8)10( f 108
5)85()85(
g
Inverse Functions
Starting with 10, we “applied” function and then “applied” function g to the result, which returned the number 10.
Inverse Functions
As further examples, check that
( ) and (3 29 2 ,3)9 f g
( ) and ( 3 5)3 5 ,55 f g
( ) and 3 38 8
2 ,2
g g
Inverse Functions
( ( )) and ( (2 2 2 ) .2) f g g f
In particular, for this pair of functions,
In fact, for any value of x,
( ( )) and ( ( )) ,x x x x f g g f
or ( )( ) and ( )( ) .x x x x f g g f
Because of this property, g is called the inverse of .
Inverse Function
Let be a one-to-one function. Then g is the inverse function of if
( )( ) x xf g for every x in the domain of g,
( )( )x xg f
andfor every x in the domain of .
EXAMPLE 1Converting from Exponential to Logarithmic Form
a. 43 64
Write each exponential equation in logarithmic form.
b. 1
2
4
1
16c. a 2 7
Solution
43a. 4 64 3log 64
4
1 2
1 1 1b. log
2 16 14
6
2 2c. 7 log 7aa
EXAMPLE 2Converting from Logarithmic Form to Exponential Form
a. log3 2435
Write each logarithmic equation in exponential form.
b. log2 5 x c. loga N x
Solution
35a. log 243 5 243 3
2b. log 5 5 2xx
c. logaxN x N a
EXAMPLE 3 Evaluating Logarithms
a. log5 25
Find the value of each of the following logarithms.
b. log2 16 c. log1 3 9
d. log7 7 e. log6 1 f. log4
1
2
Solution2
5a. log 25 25 5 or 5 5 2yyy y
42b. log 16 16 2 or 2 2 4yyy y
EXAMPLE 3 Evaluating Logarithms
Solution continued
17d. log 7 7 7 or 7 7 1y yy y
06e. log 1 1 6 or 6 6 0y yy y
1 24
1 1 1f. log 4 or 2 2
2 2 2y y yy
21 3
1c. log 9 9 or 3 3 2
3y
y
y y
EXAMPLE 4 Using the Definition of Logarithm
a. log5 x 3
Solve each equation.
b. log3
1
27y
c. logz 1000 3 22d. log 6 10 1x x
Solution
5
3
3
3a. log
5
1 1
5 125
x
x
x
EXAMPLE 4 Using the Definition of Logarithm
Solution continued
3
3
1b. log
271
327
3 3
3
y
y
y
y
3 3
3
c. log 1000
1000
1
0
3
0
1
z
z
z
z
EXAMPLE 4 Using the Definition of Logarithm
Solution continued
22
2
2
1
d. log 6 10
6 10 2 2
6 8 0
2
1
4 0
x x
x x
x x
x x
x 2 0 or x 4 0
x 2 or x 4
Rules of Logarithms with Base aIf M, N, and a are positive real numbers with a ≠ 1, and x is
any real number, then
1. loga(a) = 1 2. loga(1) = 0
3. loga(ax) = x 4.
5. loga(MN) = loga(M) + loga(N)
6. loga(M/N) = loga(M) – loga(N)
7. loga(Mx) = x · loga(M) 8. loga(1/N) = – loga(N)
Na Na )(log
Rules of Logarithms
These relationships are used to solve exponential or logarithmic equations
COMMON LOGARITHMS
1. log 10 = 1
2. log 1 = 0
3. log 10x = x4. 10log x x
The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x = log10 x. Thus,
y = log x if and only if x = 10 y.
Applying the basic properties of logarithms
NATURAL LOGARITHMS
1. ln e = 1
2. ln 1 = 0
3. log ex = x4. eln x x
The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = loge x. Thus,
y = ln x if and only if x = e y.
Applying the basic properties of logarithms
DOMAIN OF LOGARITHMIC FUNCTION
Domain of y = loga x is (0, ∞)Range of y = loga x is (–∞, ∞)
Logarithms of 0 and negative numbers are not defined.
EXAMPLE 5 Finding the Domain
3log 2f x x Find the domain of
2 0
2
x
x
Solution
Domain of a logarithmic function must be positive, that is,
The domain of f is (–∞, 2).
EXAMPLE 6 Sketching a Graph
Sketch the graph of y = log3 x.Solution by plotting points (Method 1)
Make a table of values.
EXAMPLE 6 Sketching a Graph
Solution continued
Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log3 x.
66
EXAMPLE 6 Sketching a Graph
Solution by using the inverse function (Method 2)
Graph y = f (x) = 3x.
Reflect the graph of y = 3x in the line y = x to obtain the graph of y = f –1(x) = log3 x.
PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
Domain (0, ∞) Range (–∞, ∞)
1. Domain (–∞, ∞) Range (0, ∞)
x-intercept is 1 No y-intercept
2. y-intercept is 1 No x-intercept
3. x-axis (y = 0) is the horizontal asymptote
y-axis (x = 0) is the vertical asymptote
PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
The graph is a continuous smoothcurve that passes through the points
(1, 0), and
(a, 1).
4. The graph is a continuous smooth curve that passes through the points
(0, 1), and
(1, a).
11, ,
a
1, 1 ,
a
PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
Is one-to-one, that is, logau = logav if and only if u = v.
5. Is one-to-one , that is, au = av if and only if u = v. Increasing if a > 1
Decreasing if 0 < a < 16. Increasing if a > 1 Decreasing if 0 < a < 1
EXAMPLE 7 Using Transformations
Start with the graph of f (x) = log3 x and use transformations to sketch the graph of each function.
a. f x log3 x 2
c. f x log3 x
b. f x log3 x 1
d. f x log3 x
State the domain and range and the vertical asymptote for the graph of each function.
EXAMPLE 7 Using Transformations
Solution
Shift up 2Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0
3a. l 2ogf x x
EXAMPLE 7 Using Transformations
Solution continued
Shift right 1Domain (1, ∞)Range (–∞, ∞)Vertical asymptote x = 1
3b. l 1ogf x x
EXAMPLE 7 Using Transformations
Solution continued
Reflect graph of y = log3 x in the x-axis Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0
3c. logf x x
EXAMPLE 7 Using Transformations
Solution continued
Reflect graph of y = log3 x in the y-axis Domain (∞, 0)Range (–∞, ∞)Vertical asymptote x = 0
3d. logf x x
EXAMPLE 8 Using Transformations to Sketch a Graph
Sketch the graph of y 2 log x 2 .Solution
Start with the graph of f (x) = log x.
Step 1: Replacing x with x – 2 shifts thegraph two units right.
EXAMPLE 8 Using Transformations to Sketch a Graph
Solution continued
Step 2: Multiplyingby 1 reflects the graph
Step 3: Adding 2 shifts the graph
two units up.in the x-axis.
Rules of Logarithms with Base aIf M, N, and a are positive real numbers with a ≠ 1, and x is
any real number, then
1. loga(a) = 1 2. loga(1) = 0
3. loga(ax) = x 4.
5. loga(MN) = loga(M) + loga(N)
6. loga(M/N) = loga(M) – loga(N)
7. loga(Mx) = x · loga(M) 8. loga(1/N) = – loga(N)
Na Na )(log
Rules of Logarithms
EXAMPLE 1Using Rules of Logarithms to Evaluate Expressions
5a. log yz
Given that log 5 z = 3 and log 5 y = 2, evaluate each expression.
75b. log 125y
5c. logz
y 1/30 55d. log z y
Solution 55 5a. log
3
l loog
2
5
gyz y z
EXAMPLE 1Using Rules of Logarithms to Evaluate Expressions
7 75 5 5
5 53
b. log 125 log 125 log
lo logg 5 7
3 1727
y
y y
1/2
5 555 log
2
1c. log lo logg
2
1
2 23
1
z z
yz y
y
Solution continued
EXAMPLE 1Using Rules of Logarithms to Evaluate Expressions
Solution continued
5
1/30 5 1/30 55 5 5
5
d. log log log
15
301
530
log
3
0.1 10
10.
2
1
log
z y z y
z y
EXAMPLE 2 Writing Expressions In Expanded Form
32
2 4
1a. log
2 1
x x
x
Write each expression in expanded form.
3 2 5b. ln x y z
Solution
323 42
2 2 24
3 422 2 2
2 2 2
1a. log log 1 log 2 1
2 1
log log 1 log 2 1
2log 3log 1 4log 2 1
x xx x x
x
x x x
x x x
EXAMPLE 2 Writing Expressions In Expanded Form
1/23 2 5 3 2 5
3 2 5
3 2 5
b. ln ln
1ln
21
ln ln ln21
3ln 2ln 5ln23 5
ln ln ln2 2
x y z x y z
x y z
x y z
x y z
x y z
Solution continued
EXAMPLE 3 Writing Expressions in Condensed Form
a. log3 log 4x y
Write each expression in condensed form.
21b. 2ln ln 1
2x x
2 2 2c. 2log 5 log 9 log 75
21d. ln ln 1 ln 1
3x x x
EXAMPLE 3 Writing Expressions in Condensed Form
3a. log3 log 4 log
4
xx y
y
2 2
2 2
1/22b. 1
ln ln 1 ln ln 1
ln 1
22 x x x x
x x
Solution
EXAMPLE 3 Writing Expressions in Condensed Form
2 2 2
2 2 2
2 2
2
2
2
c. 2log 5 log 9 log 75
log log log 75
log log 75
25 9log
75lo
5
g
9
25 9
3
Solution continued
EXAMPLE 3 Writing Expressions in Condensed Form
Solution continued
2
2
2
32
1d. ln 1
31
ln 13
11ln
3 1
ln ln 1
ln
1n
1
l1
x
x
x x
x
x x
x
x x
x x
CHANGE-OF-BASE FORMULA
Let a, b, and x be positive real numbers with a ≠ 1 and b ≠ 1. Then logb x can be converted to a different base as follows:
log log lnlog
log log ln
(ba base 1se ) ( ) ( )bas0 e
ab
a
x x x
e
xb b
a
b
EXAMPLE 4Using a Change of Base to Compute Logarithms
Compute log513 by changing to a. common logarithms and b. natural logarithms.
5
13lnb.
513 log
ln1.59369
5
13loga. log
log
1.5
13
69
5
93
Solution
EXAMPLE 9 Evaluating the Natural Logarithm
Evaluate each expression.
a. ln e4 b. ln1
e2.5 c. ln 3
Solution
4a. ln 4e
2.2.5
51b. ln l .5n 2e
e
Use a calculator.c. ln 31.0986123
EXAMPLE 10 Doubling Your Money
a. How long will it take to double your money if it earns 6.5% compounded continuously?
b. At what rate of return, compounded continuously, would your money double in 5 years?
Solutiona. If P is the original
amount invested, A = 2P.
It will take 11 years to double your money.
EXAMPLE 10 Doubling Your Money
Solution continued
b. Your investment will double in 5 years at the rate of 13.86%.
Solving Exponential Or Logarithmic EquationsTo solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a > 0 and a ≠ 1, and follow the guidelines.1.ax = b Solve by taking logarithms on both sides.2. Loga x = b Solve by changing to exponential form ab = x.
SOLVING AN EXPONENTIAL EQUATION
Solve 7x = 12. Give the solution to the nearest thousandth.
Solution
While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here.
SOLVING AN EXPONENTIAL EQUATION
Solve 7x = 12. Give the solution to the nearest thousandth.
Solution 7 12x
I 7 12n Inx Property of logarithms
In 7 In 12x Power of logarithms
In12In 7
x Divide by In 7.
1.277x Use a calculator.
The solution set is {1.277}.
SOLVING AN EXPONENTIAL EQUATION
Solve 32x – 1 = .4x+2 . Give the solution to the nearest thousandth.
Solution 2 1 23 .4x x
2 1 2In In 3 .4x x Take natural logarithms on both sides.
(2 1) In 3 ( 2) In .4x x Property power
2 In 3 In 3 In .4 2 In .4x x Distributive property
SOLVING AN EXPONENTIAL EQUATION
Solve 32x – 1 =.4x+2 . Give the solution to the nearest thousandth.
Solution
2 In 3 In .4 2 In .4 In 3x x Write the terms with x on one side
(2 In 3 In .4) 2 In .4 In 3x Factor out x.
2 In .4 32 In 3 .4
x
Divide by 2 In 3 – In .4.
2
2
In .4 In 3In 3 In .4
x
Power property
SOLVING AN EXPONENTIAL EQUATION
Solve 32x – 1 =.4x+2 . Give the solution to the nearest thousandth.
Solution In .16 In 3
In 9 In .4x
Apply the exponents.
In .489
In .4
x Product property; Quotient property
.236x This is approximate.
This is exact.
The solution set is { –.236}.
SOLVING BASE e EXPONENTIAL EQUATIONS
Solve the equation. Give solutions to the nearest thousandth.
Solution
a.2
200x e
2
200x e
2
In In 200x e Take natural logarithms on both sides.
2 In 200x In = x22xe
SOLVING BASE e EXPONENTIAL EQUATIONS
Solution
a.2
200x e
Square root propertyIn 200x
Remember both roots.
2.302x Use a calculator.
The solution set is { 2.302}.
Solve the equation. Give solutions to the nearest thousandth.
SOLVING BASE e EXPONENTIAL EQUATIONS
Solution
b.
Take natural logarithms on both sides.
2 1 3x e e m n m na a a
2In In 3x e
2 In In 3x e Power property
Solve the equation. Give solutions to the nearest thousandth.
2 3x e Divide by e; .m
m nn
aa
a
eee xx 3412
eee xx 3412
SOLVING BASE e EXPONENTIAL EQUATIONS
Solution
b.
2 In 3x In e = 1
1In 3
2x Multiply by – ½
.549x
The solution set is {– .549}.
Solve the equation. Give solutions to the nearest thousandth.
eee xx 3412
SOLVING A LOGARITHMIC EQUATION
Solve log(x + 6) – log(x + 2) = log x.
Solution
log( 6) log( 2) logx xx
lo6
og2
glxx
x
Quotient property
62
xx
x
Property of logarithms
6 ( 2)x x x
SOLVING A LOGARITHMIC EQUATION
Solve log(x + 6) – log(x + 2) = log x.
Solution Distributive property
26 2x x x
Standard form2 6 0x x
( 3)( 2) 0x x Factor.
3 or 2x x Zero-factor property
The proposed negative solution (x = – 3) is not in the domain of the log x in the original equation, so the only valid solution is the positive number 2, giving the solution set {2}.
SOLVING A LOGARITHMIC EQUATION
Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s).
Solution log(3 2) l 1 1og( )x x
log(3 2) log( 1) log10x x Substitute.
log[(3 2)( 1)] log10x x Product property
(3 2)( 1) 10x x Property of logarithms
SOLVING A LOGARITMIC EQUATION
Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s).
Solution 23 2 10x x Multiply.
23 12 0x x Subtract 10.
1 1 1446
x Quadratic formula
SOLVING A LOGARITMIC EQUATION
Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s).
Solution
1 1456
The number is negative, so x – 1 is negative. Therefore, log(x – 1) is not defined and this proposed solution must be discarded.
Since > 1, both 3x + 2 and x – 1 are positive and the solution set is
1 1456
1 145.
6
NEWTON’S LAW OF COOLING
Newton’s Law of Cooling states that
where T is the temperature of the object at time t, Ts is the surrounding temperature, and T0 is the value of T at t = 0.
T Ts T0 Ts e kt ,
EXAMPLE 11 McDonald’s Hot Coffee
The local McDonald’s franchise has discovered that when coffee is poured from acoffeemaker whose contents are 180ºF into a noninsulated pot, after 1 minute, the coffee cools to 165ºF if the room temperature is 72ºF. How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at 125ºF?
EXAMPLE 11 McDonald’s Hot Coffee
Use Newton’s Law of Cooling with T0 = 180 and Ts = 72 to obtain
Solution
We have T = 165 and t = 1.
1807
7 08
2
1
72
2
kt
kt
T e
T e
72 108
93
1 8
165
0
k
k
e
e
93ln
108
0.1495317
k
k
EXAMPLE 11 McDonald’s Hot Coffee
Substitute this value for k.
Solution continued
Solve for t when T = 125.
T 72 108e 0.1495317t
0.1495317
0.1495317
1 72 108
125 72
10853
ln 0.14953 7
5
8
2
110
t
t
e
e
t
1 53ln
0.1495317 108
4.76
t
t
The employee should wait about 5 minutes.
GROWTH AND DECAY MODEL
A is the quantity after time t.A0 is the initial (original) quantity (when t = 0).r is the growth or decay rate per period.t is the time elapsed from t = 0.
0rtA A e
EXAMPLE 12 Chemical Toxins in a Lake
A chemical spill deposits 60,000 cubic meters of soluble toxic waste into a large lake. If 20% of the waste is removed every year, how many years will it take to reduce the toxin to 1000 cubic meters?
Solution
In the equation A = A0ert, we need to find A0, r, and the time when A = 1000.
EXAMPLE 12 Chemical Toxins in a Lake
60,000 rtA e
1. Find A0. Initially (t = 0), we are given A0 = 60,000. So
Solution continued
2. Find r. When t = 1 year, the amount of toxin will be 80% of its initial value, or