Transcript chapter 6 notes,  

Introduction to Exponential
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Suppose you are given a number, say 2,
and you do the following to it:
2 8
Now, to get 2 back
we can take 3rd root of 8:
3
82
But, suppose you start with 3,
3
and do the following to it;
exponentiate it with base 2:
2 8
3
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
2 8
How do you get 3 back from 8?
The operation we do to 8 to get 3 back
is called logarithm,
or we take logarithm of 8 with base 2:
log 2 8  3
3
The answer to logarithm is
the exponent of the base
that gives the argument.
In this chapter we create functions using the operations of exponentiation and
logarithm, and study the behavior of such functions.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
We start by creating exponential functions,
but restrict the base to numbers (0,1) and (1, ).
The reason for this restriction is that we like to be able to
conclude that: when the bases are equals,
then the exponents are equals:
But , if 1  1 or 0  0
 we cannot conclude that 3  4!
3
4
3
4
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
We define two families of exponential functions:
f ( x)  b , b  1
examples :
x
f ( x)  2
x
f ( x)  10
f ( x)  e
1
f ( x)   
2
x
x
x
x
3
f ( x)   
2
f ( x)  b , 0  b  1
examples :
x
x
2
f ( x)   
3
x
f ( x)  (0.23)
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
x
The graphs of f ( x)  b when b  1 "look alike."
We graph an example of this family:
f ( x)  2
x
All members of this
family are increasing
functions.
Domain : (, )
Range : (0, )
when x  , 2 x  
when x  , 2  0
x
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
The graphs of f ( x)  b when 0  b  1 "look alike."
We graph an example of this family:
x
x
1
f ( x)    ,
2
x
but this is the same as f ( x )  2 ,
which is a reflection of f ( x )  2 over the y  axis !
x
College Algebra chapter 6
1
f ( x)   
2
x
Introduction to Exponential and Logarithmic Functions
All members of this
family are decreasing
functions.
x
when x  , 2  0
x
when x  , 2  
Domain : (, )
Range : (0, )
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
The value of a car can be modeled by
x
4
V ( x)  25   ,
5
where x  0 is the age of the car in years
and V ( x) is the value in thousands of dollars.
0
4
1. Find and interpret V (0).
V (0)  25    25
5
1) Since x represents the age of the car in years, x = 0 corresponds to
the car being brand new. Since V (x) is measured in thousands of
dollars, V (0) = 25 corresponds to a value of $25,000. Putting it all
together, we interpret V (0) = 25 to mean the purchase price of the car
was $25,000.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
x
4
The value of a car can be modeled by V ( x)  25   ,
5
where x  0 is the age of the car in years and
V ( x) is the value in thousands of dollars.
2. Sketch the graph of y  V ( x).
3. Find and interpret the horizontal asymptote of the graph you found in 2.
2) The horizontal asymptote
remains y = 0. Finally, we
restrict the domain to [0,∞)
to fit with the applied
domain given to us. We
have the result below.
College Algebra chapter 6
3) We see from the
graph of V that its
horizontal asymptote
is y = 0. This means
as the car gets older,
its value diminishes
to 0.
Introduction to Exponential and Logarithmic Functions
According to Newton's Law of Cooling the temperature of coffee
T (in degrees Fahrenheit)
t minutes after it is served can be modeled by
0.1t
T (t )  70  90e .
1. Find and interpret T (0).
2. Sketch the graph of y  T (t ).
3. Find and interpret the horizontal
asymptote of the graph.
T (0)  70  90e
0.1(0)
 160
College Algebra chapter 6
1) This means
that the coffee
was served at
160 degrees F.
Introduction to Exponential and Logarithmic Functions
According to Newton's Law of Cooling the temperature of coffee
T (in degrees Fahrenheit)
t minutes after it is served can be modeled by
0.1t
T (t )  70  90e .
1. Find and interpret T (0).
2. Sketch the graph of y  T (t ).
3. Find and interpret the horizontal
asymptote of the graph.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
According to Newton's Law of Cooling the temperature of coffee
T (in degrees Fahrenheit)
t minutes after it is served can be modeled by
0.1t
T (t )  70  90e .
3. Find and interpret the horizontal
asymptote of the graph.
3) The graph is approaching the horizontal line
y = 70 from above. This means that as time goes by,
the temperature of the coffee is cooling to 70◦F,
presumably room temperature.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Introduction to Logarithmic Functions
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
The inverse of the exponential function f ( x)  b
is called the base b logarithm function, and
1
is denoted f ( x)  log b ( x) .
The expression log b ( x) is read log base b of x.
For each family of exponential functions we define a family of
logarithm functions. The logarithm functions are inverses of
exponential functions.
College Algebra chapter 6
x
Introduction to Exponential and Logarithmic Functions
We define two families of logarithm functions:
f ( x)  log b x , b  1
examples :
f ( x)  log 2 x
f ( x)  log10 x  log x
f ( x)  log e x  ln x
f ( x)  log  3 2 x
f ( x)  log b x , 0Forexample:
b 1
examples :
f ( x)  log 1 2 x
f ( x)  log  2 3 x
f ( x)  log 1 3 x
College Algebra chapter 6
The inverse of f ( x )  2 x
is f 1 ( x)  log 2 x.
The inverse of f ( x )  10 x
is f 1 ( x)  log x.
The inverse of f ( x )  e x
is f 1 ( x)  ln x.
For example:
Introduction to Exponential and Logarithmic Functions
We define two families of logarithm functions:
The inverse of f ( x )  2
f ( x)  log b x , b  1
x
examples :
f ( x)  log 2 x
1
is f ( x)  log 2 x.
f ( x)  log10 x  log x
The inverse of f ( x )  10
1
x
f ( x)  log e x  ln x
f ( x)  log  3 2 x
is f ( x)  log x.
f ( x)  log b x , 0  b  1
The inverse of f ( x )  e
examples :
f ( x)  log 1 2 x
1
is f ( x)  ln x.
x
f ( x)  log  2 3 x
f ( x)  log 1 3 x
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
The graphs of logarithm functions are reflections of the graphs of
exponential functions over the line y=x.
1
f ( x)  e
Domain : ( , )
Range : (0, )
Horizontal assymptote : y  0
Goes through (0,1)
It is always increasing
x
College Algebra chapter 6
f ( x)  ln x
Domain : (0, )
Range : ( , )
Vertical assymptote : x  0
Goes through (1,0)
It is always increasing
Introduction to Exponential and Logarithmic Functions
The graphs of logarithm functions are reflections of the graphs of exponential
functions over the line y=x.
f ( x)  10
Domain : ( , )
Range : (0, )
Horizontal assymptote : y  0
Goes through (0,1)
It is always increasing
1
f ( x)  log x
Domain : (0, )
Range : ( , )
Vertical assymptote : x  0
Goes through (1,0)
It is always increasing
x
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
The graphs of logarithm functions are reflections of the graphs of
exponential functions over the line y=x.
x
1
f ( x)   
2
Domain : ( , )
Range : (0, )
Horizontal assymptote : y  0
Goes through (0,1)
It is always decreasing.
College Algebra chapter 6
1
f ( x)  log1 2 x
Domain : (0, )
Range : ( , )
Vertical assymptote : x  0
Goes through (1,0)
It is always decreasing.
Introduction to Exponential and Logarithmic Functions
y  b 

x
Take log base b of both sides
log b y  log b b  log b y  x
x
y  log b x 
Exponentiate both sides with base b
b b
b  x
Logarithm and exponentiation
with same base cancel each other.
y
logb x
y
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Simplify the following:.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Simplify the following:.
6) 2
log 2 8
7) 117
8
 log117 6
Note : 2
()
and log 2 ( ) cancel
1
1


log117 6
117
6
Note :117
College Algebra chapter 6
()
and log117 ( ) cancel
Introduction to Exponential and Logarithmic Functions
Find the domain of the following.
3  x  0  x  3  D : (,3)
x
 0  D : (,0)  (1, )
x 1
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
x 1
Let f ( x)  2  3.
1. Graph f using transformations and state the domain and range of f .
1
2. Explain why f is invertible and find a formula for f ( x).
D : (, ) R : ( 3, ) The HA y  0 is moved down 3 units.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
x 1
Let f ( x)  2  3.
1. Graph f using transformations
and state the domain and range of f .
f ( x) is invertible because
it passes the HLT.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Let f ( x)  2
2. Explain why f is invertible
1
and find a formula for f ( x).
f ( x)  2
y2
x 1
log 2 (2
x 1
3
3 x  2
y 1
y 1
3 2
y 1
 x3
)  log 2 ( x  3)  y  1  log 2 ( x  3)
1
y  log 2 ( x  3)  1  f ( x)  log 2 ( x  3)  1
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Algebraic Properties of Exponential Functions
Let f ( x)  b be an exponential function,
b  0, b  1, and let u and w be real numbers.
x
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
This is the extension of rules
with integer powers to real number powers.
2 3  2 4  2 3 4  2 7
2 2
3
5
2
3 5
Note: the exponential function turns addition into product ,
and subtraction into quotient. So, we should expect the
inverse of exponential, logarithm, do the inverse! And, as we
see next, it does.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Algebraic Properties of Logarithm Functions
Let g( x)  log b x be a logarithm function,
b  0, b  1, and let u  0 and w  0 be real numbers.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Note: the logarithm function turns multiplication into addition,
division into subtraction, and exponents into coefficients. These
properties were used to simplify operations using logarithm. Before
the invention of calculators in 1970’s students used slide rules, which
were built based on properties of logarithm.
To multiply A by B :
AB  ?  log AB  log?
log A  log B  log?
#  log?
10  10  10  ?  the answer !
#
College Algebra chapter 6
?
#
Introduction to Exponential and Logarithmic Functions
Expand the following using the properties of logarithms and simplify. Assume
when necessary that all quantities represent positive real numbers.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Expand the following using the properties of logarithms and simplify. Assume
when necessary that all quantities represent positive real numbers.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Expand the following using the properties of logarithms and simplify. Assume
when necessary that all quantities represent positive real numbers.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Expand the following using the properties of logarithms and simplify. Assume
when necessary that all quantities represent positive real numbers.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Note that the two functions f ( x)  log117 ( x 2  4)
and g ( x)  log117 ( x  2)  log117 ( x  2)
have different domains, and as a result,
are different functions.
D f : (, 2)  (2, ),
and Dg : (2, ).
In general, when using log properties to expand a logarithm,
we may be restricting the domain as we do so.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Use the properties of logarithms to write the following as a single logarithm.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Use the properties of logarithms to write the following as a single logarithm.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
Use the properties of logarithms to write the following as a single logarithm.
College Algebra chapter 6
Introduction to Exponential and Logarithmic Functions
The change of base rule :
For logarithm :
ln x log x log m x
log b x 


for x  0
ln b log b log m b
For Exponentiation:
a b
x
logb a x
b
x logb a
for all real number x
College Algebra chapter 6
Example :
Introduction to Exponential and Logarithmic Functions
ln 3 log 3 log m 3
log 2 3 


,
ln 2 log 2 log m 2
where m can be any base
Example :
x
Write 2 in base 3.
log3 2 x
2 3
3
we can calculate log 3 2,
x
( x log 3 2)
The change of base rule :
For logarithm :
ln x log x log m x
log b x 


for x  0
ln b log b log m b
For Exponentiation:
a b
x
logb a x
 b x logb a
and replace it in the exponent with a decimal.
College Algebra chapter 6
for all real number x
Introduction to Exponential and Logarithmic Functions
Use an appropriate change of base formula to convert the following expressions
to ones with the indicated base.
3  10
2
2 e
x
2
log10 3
log e 2 x
 10
e
2 log10 3
x log e 2
 10
e
x ln 2
ln 5
log 4 5 
ln 4
log x
ln x 
log e
College Algebra chapter 6
2 log 3
Introduction to Exponential and Logarithmic Functions
To use a calculator to find log b a, use the change of base
formula to change the base to 10 or e :
Example :
ln 3 log 3
log 2 3 

 1.584962501
ln 2 log 2
ln 3
log 3
log1 2 3 

 1.584962501
ln(1 2) log(1 2)
College Algebra chapter 6
Exponential Equations and Inequalities
The number e : The discovery of the constant itself is credited to Jacob
Bernoulli, who attempted to find the value of the following expression:
n
 1
1    ?
 n
when n  
Jacob Bernoulli discovered this constant by studying a
question about compound interest:
An account starts with $1.00 and pays 100 percent interest per year.
If the interest is credited once, at the end of the year, the value of
the account at year  end will be $2.00. What happens if the interest
is computed and credited more frequently during the year ?
College Algebra chapter 6
Compound interest formula:Exponential Equations and Inequalities
nt
 r
A  p 1  
 n
If APR is 100% for one year and
compounding is once per year : p  1, r  1, n  1, t  1:
(1)(1)
 1
A  11  
 $2
 1
If the interest is credited twice in the year :
(2)(1)
 1
A  11  
 $2.25
 2
If the interest is credited more often in the year :
 1
A  11  
 n
( n )(1)
 ? when n  
College Algebra chapter 6
Exponential Equations and Inequalities
An account starts with $1.00 and pays 100 percent interest per year.
If the interest is credited once, at the end of the year, the value of
the account at year  end will be $2.00. What happens if the interest
is computed and credited more frequently during the year ?
 1
A  11  
 n
( n )(1)
 ? when n  
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
College Algebra chapter 6
 1
A  11  
 n
( n )(1)
Exponential Equations and Inequalities
 ? when n  
“Bernoulli noticed that this sequence approaches a limit (the
force of interest) with larger n and, thus, smaller
compounding intervals. Compounding weekly (n = 52) yields
$2.692597..., while compounding daily (n = 365) yields
$2.714567..., just two cents more. The limit as n grows large
is the number that came to be known as e; with continuous
compounding, the account value will reach $2.7182818....”
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
College Algebra chapter 6
Exponential Equations and Inequalities
In this section we will develop techniques for solving equations involving
exponential functions.
To solve an equation such as
2  32
make the bases the same on both sides,
then set the exponents equal to each other:
x
2  2  x  5.
x
5
College Algebra chapter 6
Exponential Equations and Inequalities
In this section we will develop techniques for solving equations involving
exponential functions.
But , we cannot use this method to solve
2  33!
x
To solve exponential equations take
ln (or log any base) of both sides :
ln 33
ln(2 )  ln(33)  x ln 2  ln 33  x 
 5.044394119
ln 2
x
College Algebra chapter 6
Exponential Equations and Inequalities
Solve the following equations. Check your answer graphically using a calculator.
2  16
3x
2 2
3x
(1 x )
(4 4 x )
4 (1 x )
 2  (2 )
3x
2 2
3x
4(1 x )
 3 x  4  4 x  7 x  4  x  4 7  0.5714
2000
2000  1000(3 ) 
 30.1t  2  30.1t  ln 2  ln 30.1t
1000
ln 2
ln 2  0.1t ln 3  t 
 6.3093
(0.1)ln 3
0.1t
College Algebra chapter 6
Exponential Equations and Inequalities
Solve the following equations. Check your answer graphically using a calculator.
9(3 )  7  3  3  7
x
2x
x2
2
x
x2
2x
 3  7  ln 3  ln 7
 ( x  2)ln 3  2 x ln 7
 x(ln 3)  2(ln 3)  2 x(ln 7)
 2(ln 3)  2 x(ln 7)  x(ln 3)
 2(ln 3)  x(2(ln 7)  (ln 3))
2(ln 3)
x
 0.7866
2(ln 7)  (ln 3)
College Algebra chapter 6
2x
2x
Exponential Equations and Inequalities
Solve the following inequalities. Check your answer graphically using a calculator.
1) Turn into an equation, and solve:
x
x
x
e
e
e
3 x
3 x
3 0
x
e 4
e 4
e 4
x
e
x
x
(e  4) x
 3(e  4)  0
e 4
x
x
 e  3e  12  0
2e  12  e  6  ln e  ln 6  x  ln 6
x
x
x
Also, e  4  0  e  4  x  ln 4
x
x
College Algebra chapter 6
Exponential Equations and Inequalities
Solve the following inequalities. Check your answer graphically using a calculator.
x  ln 6
x  ln 4
x
e
2) Sign of x
 3:
e 4
     ln 4        ln 6     
The solution : (,ln 4)  (ln 6, ).
The solution : (,ln 4)  (ln 6, )
College Algebra chapter 6
Logarithmic Equations and Inequalities
In this section we will develop techniques for solving equations involving
logarithmic functions.
To solve an equation such as log 3 x  log 3 5 ,
we set the arguments equal each other
because the bases are equals:
log 3 x  log 3 5  x  5
But we cannot use this method to solve: log 2 x  5!
To solve logarithmic equations such as log 2 x  5,
we convert to an exponential equation:
log 2 x  5 
Exponentiate both sides with base 2
2log2 x  25  x  25  32.
College Algebra chapter 6
Logarithmic Equations and Inequalities
Solve the following equation. Check your
solutions graphically using a calculator.
log117 (1  3 x)  log117 ( x  3) 

2
Bases are the same
(1  3 x)  ( x 2  3)
x  3 x  4  0  ( x  4)( x  1)  0
 x  1, x  4, only  4 checks
Use change of base formula to write the eqaution as
2
ln(1  3 x) ln( x  3)

,
ln117
ln117
and use a calculator to check the solutions :
the solution set  {4}
2
College Algebra chapter 6
Logarithmic Equations and Inequalities
Solve the following equation. Check your solutions graphically using a calculator.
only  4 checks
College Algebra chapter 6
Logarithmic Equations and Inequalities
Solve the following inequality. Check your answer graphically using a calculator.
1
Turn
 1  0 into an equation, and solve:
ln x  1
1
1
 1  (ln x  1)
 (ln x  1)1
ln x  1
ln x  1
0
 1  ln x  1  ln x  0  x  e  1
Also, ln x  1  0
1
1
 ln x  1  x  e  x 
e
College Algebra chapter 6
Logarithmic Equations and Inequalities
1
0             1      
e
1
 1.
Use a calculator to find the sign of
ln x  1
From our sign diagram,
1
we find the solution to be (0, )  [1, ).
e
College Algebra chapter 6
Applications of Exponential
and Logarithmic Functions
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Simple Interest: The amount of interest I
accrued at an annual rate r on an
investment P after t years is
I  Pr t
The amount A in the account after t years is given by
A  P  I  P  Pr t  P (1  rt )
College Algebra chapter 6
A  P (1  rt )
Applications of Exponential and Logarithmic Functions
Suppose you have $100 to invest at your local bank
and they are offering 5% annual percentage interest
rate. This means that after one year, the bank will pay
you 5% of that $100, or $100(0.05) = $5 in interest, so
you now have $105.1.
A  100(1  0.05(1))  105
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Compounded Interest:
If an initial principal P is invested at an annual rate r
and the interest is compounded n times per year,
the amount A in the account after t years is
 r
A(t )  P 1  
 n
nt
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Example: Suppose $2000 is invested in an account which offers
7.125% compounded monthly.
1. Express the amount A in the account as a function of the
term of the investment t in years.
P  2000, r  0.07125, n  12 (compounded monthly )
12 t
 0.07125 
A(t )  2000 1 

12 

12 t
A(t )  2000(1.0059375)
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Example: Suppose $2000 is invested in an account which offers
7.125% compounded monthly.
2. How much is in the account after 5 years?
Since t represents the length of
the investment in years,
we substitute t = 5 into A(t) to find
A(5)  2000(1.0059375)
12(5)
College Algebra chapter 6
 2852.92
Applications of Exponential and Logarithmic Functions
Example: Suppose $2000 is invested in an account which offers
7.125% compounded monthly.
3. How long will it take for the initial investment to double?
Our initial investment is $2000,
so to find the time it takes this to double, we
need to find t when A(t) = 4000.
 4000  (1.0059375)  2
ln 2
12 t
ln(1.0059375)  ln 2  t 
 9.75
12ln(1.0059375)
It takes 9 years and 9 months for the investment to double.
12 t
12 t
2000(1.0059375)
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Suppose $2000 is invested in an account which offers 7.125% compounded monthly.
4. Find and interpret the average rate of change of the amount in the account
from the end of the fourth year to the end of the fifth year, and from the end of
the thirty-fourth year to the end of the thirty-fifth year.
To find the average rate of change of A from
the end of the fourth year to the end of the
A(5)  A(4)
fifth year, we compute
 195.63.
54
Similarly, the average rate of change of A from
the end of the thirty-fourth year to the end of
A(35)  A(34)
the thirty-fifth year is
 1648.21.
35  34
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Suppose $2000 is invested in an account which offers 7.125% compounded monthly.
4. Find and interpret the average rate of change of the amount in the account
from the end of the fourth year to the end of the fifth year, and from the end of
the thirty-fourth year to the end of the thirty-fifth year.
This means that the value of the investment
is increasing at a rate of approximately
$195.63 per year between the end of the
fourth and fifth years, while that rate jumps
to $1648.21 per year between the end of the
thirty-fourth and thirty-fifth years. So, not
only is it true that the longer you wait, the
more money you have, but also the longer
you wait, the faster the money increases.
College Algebra chapter 6
How to calculate the average
with a calculator:
Applications of Exponential and Logarithmic Functions
Continuously Compounded Interest: If an initial principal P is invested at an annual rate r and
the interest is compounded continuously, the amount A in the account after t years is
rt
A(t )  Pe
Suppose $2000 is invested in an account which offers 7.125% with
continuous compounding. How much is accumulated over 35 years?
Compare monthly compounding with continuously compounding
Monthly compounding yields
A(35)  2000(1.0059375)
 $24,035.28,
whereas continuously compounding gives
12(35)
A(35)  2000e
=$24, 213.18,
a difference of less than 1%.
0.07125(35)
College Algebra chapter 6
 r
A(t )  P 1  
 n
nt
Applications of Exponential and Logarithmic Functions
Uninhibited Growth :
If a population increases according to The Law of Uninhibited
Growth, the number of organisms N at time t is given by the formula
N (t )  N 0e ,
kt
where N (0)  N 0 (" N nought ")
is the initial number of organisms
and k  0 is the constant of proportionality
which satisfies the equation.
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
In order to perform arthrosclerosis research, epithelial cells are
harvested from discarded umbilical tissue and grown in the
laboratory. A technician observes that a culture of twelve thousand
cells grows to five million cells in one week. Assuming that the cells
follow The Law of Uninhibited Growth, find a formula for the number
of cells, N, in thousands, after t days.
N 0 12 in thousands
N (t )  N 0e kt 

N (t )  12e 
kt
we know that in one week we have 5,000,000 cells ( or 5000 thousnads )
we need to find k
N (7)  5000 
1  1250  The formula becomes
12e  5000  k  ln 
 
7  3 
7k
N (t )  12e
1  1250 
t ln 

7  3 
College Algebra chapter 6
Interesting web site
on world population
Applications of Exponential and Logarithmic Functions
Radioactive Decay:
The amount of a radioactive element A at time t
is given by the formula
A(t )  A0e ,
kt
A(0)  A0 is the initial amount of the element
and k  0 is the constant of proportionality
which satisfies the equation.
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid
is functioning. Suppose the decay of Iodine-131 follows the model given above, and that
the half-life (the time it takes for half of the substance to decay) of Iodine-131 is approximately
8 days. If 5 grams of Iodine-131 is present initially, find a function which gives the amount of
Iodine-131, A, in grams, t days later.
A0 5 grams
 A(t )  5e 
kt
to find k we use the half life , 8 days
A (8)  2.5
A(8)  (0.5)5  2.5 
 5e8 k  2.5

divide by 5 and take ln of both sides , and solve for k
1  1  sub for k in the formula
k  ln   

8 2
A(t )  5e
1 1
ln  t
8 2
1
ln    ln 2
2
 A(t )  5e
1
 t ln(2)
8
College Algebra chapter 6
Fukushima Radiation To Reach West Coast
Iodine-131causes cancer!
Applications of Exponential and Logarithmic Functions
Logistic Growth:
If a population behaves according to the assumptions
of logistic growth, the number of organisms N at time t
is given by the equation
L
N (t ) 
,
 kLt
1  Ce
where N (0)  N 0 is the initial population,
L is the limiting population,
C is a measure of how much room there is to grow
L
given by C 
1 ,
N0
and k  0 is the constant of proportinality
which satisfies the equation.
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Use your calculator to fit a logistic model to these data, using x = 0 to
represent the year 1860.
STAT,CALC
STAT,Edit
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Use your calculator to fit a logistic model to these data, using x = 0 to
represent the year 1860.
Y=, VARS, STATISTICS, EQ, Reg
This is the model.
College Algebra chapter 6
Applications of Exponential and Logarithmic Functions
Use your calculator to fit a logistic model to these data, using x = 0 to
represent the year 1860.
ZOOM 9
College Algebra chapter 6
World Population Growth. In 2009, the world population was 6.8 billion.
The exponential growth rate was 1.13% per year.
a) Find the exponential growth function.
a ) P0  6.8, k  0.0113
P (t )  P0e  P (t )  6.8e
kt
0.0113t
P (t ) is in billions,
t is in years after 2009
College Algebra chapter 6
World Population Growth. In 2009, the world population was 6.8 billion.
The exponential growth rate was 1.13% per year.
b) Estimate the population of the world in 2012 and in 2020.
b) 2012 : 2012  2009  3, t  3
P (3)  6.8e
2020 : t  11
0.00113(3)
P (11)  6.8e
 7 billion
0.00113(11)
 7.7 billion
College Algebra chapter 6
World Population Growth. In 2009, the world population was 6.8 billion.
The exponential growth rate was 1.13% per year.
c) When will the world population be 8 billion?
8
0.0113t
c) 8  6.8e

e
6.8
8
8
0.0113t
ln
 ln e
 ln
 0.0113t
6.8
6.8
8
ln
t  6.8  t  14.4 years
0.0113
0.0113t
College Algebra chapter 6
World Population Growth. In 2009, the world population was 6.8 billion.
The exponential growth rate was 1.13% per year.
d) Find the doubling time.
d ) The doubling time
8
ln
6.8
T
 t  14.4 years
0.0113
College Algebra chapter 6
Carbon Dating. In 1970, Amos Flora of Flora, Indiana, discovered teeth
and jawbones while dredging a creek. Scientists discovered that the
bones were from a mastodon and had lost 77.2% of their carbon-14.
How old were the bones at the time they were discovered?
We first find k when the half  life,
T, is 5750 years :
ln 2
ln 2
k
k 
 k  0.00012
T
5750
0.00012 t
P (t )  P0e
College Algebra chapter 6
Carbon Dating. In 1970, Amos Flora of Flora, Indiana, discovered teeth
and jawbones while dredging a creek. Scientists discovered that the
bones were from a mastodon and had lost 77.2% of their carbon-14.
How old were the bones at the time they were discovered?
The bones lost 77.2% of their
carbon  14, so 22.8% of P0 remains.
0.228 P0  P0e
0.00012 t
 0.228  e
0.00012 t
ln 0.228
ln 0.228  ln e
t 
 12.320
0.00012
The bones are about 12,32 years old
0.00012 t
College Algebra chapter 6