Transcript Exponential and Logarithmic Functions

Exponential and
Logarithmic Functions
Exponential Functions & Their Graphs
Logarithmic Functions & Their Graphs
Properties of Logarithms
Exponential and Logarithmic Equations
Exponential and Logarithmic Models
Exponential Functions
Definition of Exponential Functions
The exponential function f with base a is denoted by
f x   a x
where a > 0, a ≠ 1, and x is any real number.
Example 1
Evaluating Exponential Expressions
Use a calculator to evaluate each expression.
a. 2 3.1
b. 2
Rounded to nearest ten thousandth
a. 2 3.1  0.1166

b. 2   0.1133
Graphs of Exponential Functions
The graphs of all exponential functions have similar characteristics.
Graphs of y = ax
Example 2
Create a table
x
-3
-2
-1
0
1
2
2x
4x
Graph of y = 2x
Graph of y = 4x
Combined Graphs
Graphs of Exponential Functions
The graphs of all exponential functions have similar characteristics.
Graphs of y = a-
Example 2
x
Create a table
x
-3
-2
-1
0
1
2
2-x
4-x
Graph of y = 2-x
Graph of y = 4-x
Combined Graphs
Comparing Exponential Functions
Basic Characteristics of Exponential Functions
Function Y = ax, a >1
Y = a-x, a >1
Domain: (- ∞, ∞)
(- ∞, ∞)
Range: (0, ∞)
y-Intercept: (0, 1)
(0, ∞)
(0,1)
Increasing/Decreasing Increasing
Decreasing
Horizontal Asymptote y = 0
y=0
Continuity: Continuous
Graph of y = ax
Continuous
Graph of y = a-x
Shifting Exponential Functions
Each of the following graphs represents a transformation of the graph of f(x) = 3x
f  x   3 x 1
Because g(x) = 3x+1 = f(x+1) ,
the graph of g can be obtained by
shifting the graph of f one unit to
the left.
f x   3 x  2
Because g(x) = 3x - 2 = f(x) - 2 ,
the graph of g can be obtained by
shifting the graph of f one unit to
the left.
Shifting Exponential Functions
Each of the following graphs represents a transformation of the graph of f(x) = 3x
f x   3 x
Because g(x) = - 3x = - f(x) , the
graph of g can be obtained by
shifting the graph of f one unit to
the left.
f x   3 x
Because g(x) = 3x - 2 = f(x) - 2 ,
the graph of g can be obtained by
shifting the graph of f one unit to
the left.
The Natural Base e
In many applications the most convenient choice for a base is the irrational
number e. e ≈ 2.71828. . .
This number is known as the natural base. The function f(x) = ex is called
the natural exponential function.
Example 3
Evaluating the Natural
Exponential Function
Graph of the Natural Base
Use a calculator to evaluate each expression.
2
a. e
b.
e 1
e1
d.
e2
c.
Rounded to nearest ten thousandth
a. e 2  0.1353
b. e 1  0.3679
c. e1  2.7183
d. e 2  7.3891
Graphing Exponential Functions
Example 4
Graphing natural Exponential Functions
Graph each natural exponential function
1
2
b. g x   e 0.58 x
a. f x   2e 0.24 x
Create a Table
X
-3
-2
-1
0
1
2
3
f(x)
g(x)
Graph of f(x)
Graph of g(x)
Applications of Exponential Functions
One of the most familiar examples of exponential growth is that of an investment
earning continuously compounded interest.
Formulas for Compound Interest
After t years the balance A in an account with principal P and annual
interest rate r (in decimal form) is given by the following formulas.
nt
r

1. A  P 1  
 n
rt
2. A  Pe
Example 5
Compounding n times and Continuously
A total of $12,000 is invested at an interest rate or 9%. Find the balance
after 5 years if it is compounded quarterly and continuously.
Quarterly:
nt
4 5 
r
.
09




Formula: A  P  1   Substitution: A  12000 1 
 Solution: A  18726.11

n
Continuously:
Formula: A  Pe rt

4 
Substitution: A  12000e.09 5 
Solution: A  18819.75
Note that continuous compounding yields $93.64 more than quarterly.
Applications of Exponential Functions
Example 6
Radioactive Decay
In 1986, a nuclear reactor accident occurred in Chernobyl in what was then the
Soviet Union. The explosion spread radioactive chemicals over hundreds of square
miles, and the government evacuated the city and surrounding areas. To see why
the city is now uninhabited, consider the following model.
P  10e 0.00002845t
This model represents the amount of plutonium that remains of the initial 10
pounds after t years.
How many grams of plutonium will remain after 100 years?
P  10e 0.00002845t
P  10e 0.00002845100 
P  9.997
How many grams of plutonium will remain after 10,000 years?
P  10e 0.00002845t
P  10e 0.0000284510000 
P  .581
From the graph you can see that Plutonium has a half life of 24,360 years.
Logarithmic Functions
Previously we have studied functions and their inverses. During this study we
realized that if a function has the property were no horizontal line intersect the
graph more than once the function has an inverse.
Looking at the graph of f(x) = ax we notice that f(x) has an inverse.
Definition of Logarithmic Function
For x>0 and a≠1,
y  log a x if and only if x  a y
The function given by f  x   log a x is called the logarithmic function
with base a.
When evaluating logarithms, remember that a logarithm is an exponent.
This means that logax is the exponent to which a must be raised to
obtain x.
For instance, log28=3 because 2 raised to the 3 power is 8.
Evaluating Logarithms
Example 7
Expression
Evaluating Logarithms
Value
Justification
a. log 2 32
5
25  32
b. log 3 27
3
33  27
c. log 4 2
½
1
100
e. log 3 1
d. log 10
f. log 2 2
41 2  2
0
1
10 
100
30  1
1
21  2
-2
2
Evaluating Logarithms
Example 7
Evaluating Logarithms on the Calculator
Use the calculator to evaluate each expression.
Expression
Key Strokes
a. log10 10
b. 2 log10 2.5
c.
LOG
2 X
log10  2 
LOG
10
Display
ENTER
LOG 2.5 ENTER
(-)
2
Properties of Logarithms
1. log a 1  0
because
2. log a a  1
because
a0  1
a1  a
3. log a a x  x
because
ax  ax
4. log a x  log a y
then
x y
ENTER
1
0.7958800
ERROR
Graphing Logarithmic Functions
Example 8
Graphing a Logarithmic Function
In the same coordinate plane graph the following two functions.
x
a. f x   2
x
2x
log10x
-2
b.
-1
0
1
2
3
Natural Logarithmic Function
The Natural Logarithmic Function
The function defined by log e x  ln x
where , x  0
is called the natural logarithmic function.
Properties of Natural Logarithms
1.
2.
3.
4.
ln 1 = 0
ln e = 1
ln ex = x
If ln x = ln y, then x = y
Example 9
Example
Using Properties of Natural Logarithms
Solution
Property
a. ln 1
ln e 1  1
Property 3
b. lne2
2
Property 3
c. lne 0
0
Property 1
d. 2ln e
2
Property 2
e
Domains
of
Logarithmic
Functions
Example 10
Finding the Domains of Logarithmic Functions
Find the domain of each function
a. f(x) = ln (x – 2)
b. g(x) = ln (2 – x)
c. Ln x2
a) Because ln (x – 2) is defined only if x – 2 > 0, it follows that the
domain of f is (2, ∞).
b) Because ln (2 – x) is defined only if 2 – x > 0, it follows that the
domain of g is (- ∞, 2).
c) Because ln x2 is defined only if x2 > , it follows that the domain of h is
all real numbers except x = 0
Application of Natural Logarithms
Example 11
Human Memory Models
Students participating in a psychological experiment attended several lectures
on a subject and were given an exam. Every month for a year after the exam,
the students were retested to see how much of the material they
remembered. The average scores for the group are given by the human
memory model.
f t   75  6 lnt  1
0  t  12
where t is time in months.
a. What was the original average score? c. What was the average score after six
months?
f 0   75  6 ln0  1
f 0   75  6 ln1
f 0   75
b. What was the average score
after two months?
f 2  75  6 ln2  1
f 2  75  6 ln 3
f 2  68.4
f 6   75  6 ln6  1
f 6   75  6 ln 7
f 6   63.3
Change of Base
Change of Base Formula
Let a, b, and x be positive real numbers such that a ≠ 1 and b ≠ 1.
log x
Then log a x  b
log b a
Example 12
Changing Bases Using Common Logarithms
Change each logarithmic function to base 10 and evaluate to
nearest ten thousandth.
Example
a. log 4 30
b. log 2 14
Change of Base
log 10 30
log 10 4
log 10 14
log 10 2
Evaluate
 2.4534
 3.8074
Properties of Logarithms
Properties of Logarithms
Let a be a positive number such that a ≠ 1, and let n be a real number.
If u and v are positive real numbers, the following properties are true.
1. log a uv   log a u  log a v
u
v 
2. log a    log a u  log a v
3.
log a u n  n log a u
Example 13
1.
lnuv   ln u  ln v
2.
u
ln   ln u  ln v
v 
3.
ln u n  n ln u
Using the Properties of Logarithms
Write the logarithm in terms of ln 2 and ln 3
a.
ln 6
b. ln
2
27
ln2  3
ln 2  ln 3
ln 2  ln 27
ln 2  ln 33
ln 2  3ln 3
Using Properties of Logarithms
Example 14
Using Properties of Logarithms
1
Use the properties of logarithms to verify that  ln  ln 2
2
Change one side to match the other.
1
Original Statement
 ln  ln 2
2
Law of Negative Exponents
 ln 2 1  ln 2
 
  1ln 2  ln 2
ln un = n ln u
Simplify
ln 2  ln 2
Example 15
Rewrite Each Logarithm In Expanded Form
log 10 5x 3 y
log 10 5  log 10 x 3  log 10 y
log 10 5  3log 10 x  log 10 y
3x  5
7
ln 3x  5  ln 7
1
ln3x  5  ln 7
2
ln
Original Statement
Product Rule
Power Rule
Original Statement
Quotient Rule
Power Rule
Using Properties of Logarithms
Example 16
Rewrite each Logarithmic Expression in Condensed Form
1
Original Statement
log 10 x  3log 10 x  1
2
Power Rule Of Logarithms
log 10 x 1 2  log 10 x  13
log 10

x x  13
Example 17

Product Rule of Logarithms
Rewrite Each Logarithm In condensed Form
2lnx  2  ln x
Original Statement
lnx  22  ln x
Product Rule
ln
x  22
x
Power Rule
Exponential and Logarithmic Equations
Solving Exponential Equations
1. Isolate the exponential expression
2. Take the logarithm of both sides
3. Solve for the variable
Example 18
Solve e x  72
e x  72
 
ln e x  ln72
x  ln 72
x  4.277
Solving Logarithmic Equations
1. Rewrite the equation in exponential
form
2. Solve for the variable
Solving an Exponential Equation
Original Equation
Take logarithm of both sides
Inverse Property
Evaluate to the thousandths place
Example 19
Solving an Exponential Equation
x
Solve e  5  60
e x  5  60
e x  55
 
ln e x  ln55
Original Equation
Isolate the exponential expression
Take the logarithm of both sides
x  ln 55
Inverse Property
x  4.007
Evaluate to the thousandths place
Exponential and Logarithmic Equations
Examples 20 & 21
2x
Solve 4e  5
4e  5
4
e2 x 
5
4
lne 2x   ln 
5
4
2 x  ln
5
2x
e 2 x  3e x  2  0
e   3e  2  0
e  2e  1  0
x
Original Equation
Isolate the exponential
expression
1 4
x  ln
2 5
x  0.112
Take logarithm of both
sides
Inverse Property
e 2 x  3e x  2  0
Solve
x 2
Solving an Logarithmic Equation
x
x
e 2  0  e 1  0
ex  2  ex  1
x  ln2  x  ln1
x  .693  x  0
Original Equation
Quadratic Form
Factor
Set factors equal to zero
Isolate exponential expression
Take logarithm of both sides
Evaluate to nearest thousandths
Solve
Evaluate to nearest
thousandths
Exponential and Logarithmic Equations
Example 20
Solving an Logarithmic Equation
Solve 2ln 3x  4
2ln 3x  4
Original Equation
Isolate the Natural Logarithm
ln 3x  2
eln3 x  e 2
Exponentiate both sides
Inverse Property
3x  e 2
2
e
Solve
x
3
x  2.463
Evaluate to the thousandths place
Solve ln x  lnx  1  1
ln x  lnx  1  1
 x 
ln
 1
 x 1
x
 e1
x 1
x  ex  e
x  ex  e
x 1  e   e
e
x
e 1
Original Equation
Quotient Rule for Logarithms
Exponentiate both sides
Cross Multiply
Subtract ex from both sides
Factor
Divide both sides by 1 - e
Applications
Example 21
Doubling an Investment
You have deposited $500 in an account that pays 6.75% interest, compounded
continuously. How long will it take your money to double ? How long for it to
triple?
Double Investment
Triple Investment
rt
Formula
A  Pe
A  Pe rt
Formula
1000  500e.0675t Substitute
1500  500e.0675t Substitute
Isolate
.0675t
Isolate
2  e.0675t
ln2  ln e.0675t
ln2  .0675t
ln2
t
.0675
10.27  t
Take Logarithm
Inverse Property
Solve
Simplify
3e
ln 3  ln e.0675t
ln 3  .0675t
ln 3
t
.0675
16.28  t
Take Logarithm
Inverse Property
Solve
Simplify
Application
Example 22
Consumer Price Index for Sugar
From 1970 to 1973, the consumer Price Index ( CPI ) value y for a fixed amount
of sugar for the year t can be modeled by the equation
y  169.8  86.8 lnt
where t = 10 represent 1970. During which year did the price of sugar reach
four times its 1970 price of 30.5 on the CPI?
y  169.8  86.8 lnt
122  169.8  86.8 lnt
291.8  86.8 lnt
Formula
Substitute
Isolate
lnt  3.62 Take Logarithm
elnt  e 3.362
t  29
Inverse Property
Solve
Since t = 0 represents 1970, the price of sugar reached 4 times its1079 price in
1988.