Transcript Document

Exponential and Logarithmic Functions
5
• Exponential Functions
• Logarithmic Functions
• Compound Interest
• Differentiation of Exponential Functions
• Differentiation of Logarithmic Functions
• Exponential Functions as Mathematical Models
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Exponential Function
An exponential function with base b and exponent x
is defined by
f ( x)  b
Ex. f ( x)  3x
x
y
1
1
0
1
2
1
3
9
y
x
b  0, b  1
y  f ( x)
Domain: All reals
Range: y > 0
3
(0,1)
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Laws of Exponents
Law
Example
1. b  b  b
x
b
x y
2. y  b
b
x
y
 
3. b
x
y
b
x y
xy
4.  ab   a b
x
x x
x
x
a
a
5.    x
b
b
2 2  2  2  8
12
5
123
9
5
5
3
5
6
1
1/ 3
6 / 3
2
8
8
8 
64
1/ 2
5/ 2
6/ 2
3
 
 2m
3
 2 m  8m
1/ 3
 8 
 
 27 
3
3
3
81/ 3
2
 1/ 3 
3
27
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Properties of the Exponential
Function
y  f ( x)  b x
b  0, b  1
1. The domain is  , .
2. The range is (0,  ).
3. It passes through (0, 1).
4. It is continuous everywhere.
5. If b > 1 it is increasing on  ,  .
If b < 1 it is decreasing on  ,  .
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Examples
Ex. Simplify the expression
3x y 
2 1/ 2
x3 y 7
4
34 x8 y 2 81x5
 3 7  5
y
x y
Ex. Solve the equation
43 x 1  24 x 2
23 x1
4 x 2
2
2
26 x2  24 x2
6x  2  4x  2
2 x  4
x  2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Logarithms
The logarithm of x to the base b is defined by
y  logb x if and only if x  b
Ex. log 3 81  4;
log 7 1  0;
log1/ 3 9  2;
log 5 5  1;

7
y
 x  0

34  81
0

1
  1 -2

    81 
 3 




51  5

Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Examples
Ex. Solve each equation
a. log2 x  5
x  2  32
5
b. log 27 3  x
3  27 x
3  33 x
1  3x
1
x
3

am  an  m  n

Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Laws of Logarithms
1. l og b mn  logb m  logb n
m
2. logb    logb m  logb n
n
3. logb mn  n logb m
4. logb 1  0
5. logb b  1
Notation:
Common Logarithm log x  log10 x
Natural Logarithm ln x  log e x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Use the laws of logarithms to simplify the
expression:
7
25x y
log5
z
 log5 25  log5 x7  log5 y  log5 z1/ 2
1
 2  7 log 5 x  log 5 y  log 5 z
2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Logarithmic Function
The logarithmic function of x to the base b is
defined by
f ( x)  logb x
b  0, b  1
Properties:
1.
2.
3.
4.
5.
Domain: (0, )
Range:  , 
x-intercept: (1, 0)
Continuous on (0,)
Increasing on (0, ) if b > 1
Decreasing on (0,  ) if b < 1
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Graphs of Logarithmic Functions
Ex.
f ( x)  log3 x
f ( x)  log1/ 3 x
1
y 
 3
y 3
y
x
(0, 1)
x
y
(0, 1)
(1,0)
x
y  log3 x
(1,0)
x
y  log1/ 3 x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
e
e
x
ln x
ln e  x
x
x
and ln x
 x  0
 for any real number x 
1 2 x 1
Ex. Solve e
 10
3
e2 x1  30
2 x  1  ln(30)
Apply ln to both sides.
ln(30)  1
x
 1.2
2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
A normal child’s systolic blood pressure may be
approximated by the function p( x)  m(ln x)  b
where p(x) is measured in millimeters of mercury, x is
measured in pounds, and m and b are constants. Given
that m = 19.4 and b = 18, determine the systolic blood
pressure of a child who weighs 92 lb.
Since m  19.4, x  92, and b  18
we have p(92)  19.4(ln 92)  18
 105.72
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Compound Interest Formula
r

A  P 1  
 m
mt
A = The accumulated amount after mt periods
P = Principal
r = Nominal interest rate per year
m = Number of periods/year
t = Number of years
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Find the accumulated amount of money after 5
years if $4300 is invested at 6% per year
compounded quarterly.
r

A  P 1  
 m
mt
 .06 
A  4300 1 

4 

4(5)
= $5791.48
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
How long will it take an investment of $10,000 to
grow to $15,000 if it earns an interest at the rate of
12% / year compounded quarterly?
r 

A  P 1  
m

mt
4t
0.12 

 15, 000  10, 000 1 
 
4 

1.5  1.03  ln(1.03) 4t  ln1.5
4t
4t ln1.03  ln1.5
Taking ln both sides
logb m n  n logb m
ln1.5
t
 3.43 years
4 ln1.03
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Effective Rate of Interest
m
r

reff  1    1
 m
reff = Effective rate of interest
r = Nominal interest rate/year
m = number of conversion periods/year
Ex. Find the effective rate of interest corresponding to
a nominal rate of 6.5% per year, compounded monthly.
m
reff
12
r
 .065 

 1    1   1 
  1  .06697
12 

 m
It is about 6.7% per year.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Present Value Formula for
Compound Interest
r

P  A 1  
 m
 mt
A = The accumulated amount after mt periods
P = Principal
r = Nominal interest rate/year
m = Number of periods/year
t = Number of years
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Find the present value of $4800 due in 6 years
at an interest rate of 9% per year compounded
monthly.
 mt
r

P  A 1  
 m
 .09 
P  4800 1 

12


12(6)
 $2802.83
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Continuous Compound Interest
Formula
A  Pe
rt
A = The accumulated amount after t years
P = Principal
r = Nominal interest rate per year
t = Number of years
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Find the accumulated amount of money after
25 years if $7500 is invested at 12% per year
compounded continuously.
A  Pe
rt
A  7500e0.12(25)
 $150,641.53
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Differentiation of Exponential
Functions
Derivative of Exponential Function
d x
e   e x
dx  
Chain Rule for Exponential Functions
If f (x) is a differentiable function, then
d
e f ( x )   e f ( x )  f ( x)

dx 
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Examples
35 x
f
(
x
)

e
.
Find the derivative of
f ( x)  e
3 5 x
d
3  5x 
dx
 5e35 x
Find the relative extrema of f ( x)  x e .
4 4x
3 4x
4 4x

f ( x)  4x e  4x e
 4 x3e4 x 1  x 
Relative Min. f (0) = 0
1
Relative Max. f (-1) = 4
f
–
+
-1
+
0
x
e
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Differentiation of Logarithmic
Functions
Derivative of Exponential Function
d
1
ln x  
dx
x
 x  0
Chain Rule for Exponential Functions
If f (x) is a differentiable function, then
d
f ( x)
ln f ( x) 
dx
f ( x)
 f ( x)  0
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Examples


Find the derivative of f ( x)  ln 2 x  1 .
d  2 
2 x  1
4x

dx

f ( x) 
2
2
2
x
1
2x 1
2
Find an equation of the tangent line to the graph of
f ( x)  2x  ln x at 1,2.
1
f ( x)  2 
x
f (1)  3
Slope:
y  2  3( x  1)
Equation:
y  3x  1
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Logarithmic Differentiation
1. Take the Natural Logarithm on both sides
of the equation and use the properties of
logarithms to write as a sum of simpler
terms.
2. Differentiate both sides of the equation with
respect to x.
dy
3. Solve the resulting equation for
.
dx
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Examples
Use logarithmic differentiation to find the derivative
5
of
y  3x  2  9 x  1
ln y  ln


3x  2 9 x 1
5


Apply ln
ln y  ln 3x  2  ln  9 x  1
1
ln y  ln  3 x  2   5ln  9 x  1
2
1 dy
3
5(9)


y dx 2  3x  2  9 x  1
5
Properties
of ln
Differentiate
dy
3
45  Solve
5
 3x  2  9 x  1 



dx
 2  3x  2  9 x  1 
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Exponential Growth/Decay
Models
A quantity Q whose rate of growth/decay at any
time t is directly proportional to the amount present
at time t can be modeled by:
Growth
Decay
Q(t )  Q0e
kt
Q(t )  Q0e
 kt
0  t  
0  t  
Q0 is the initial quantity
k is the growth/decay constant
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
A certain bacteria culture experiences exponential
growth. If the bacteria numbered 20 originally and
after 4 hours there were 120, find the number of
bacteria present after 6 hours.
Q0  20 so Q(t )  20ekt
k4
120  20e
ln 6
k
 0.4479
4
Q(6)  20e
0.4479(6)
 294
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Learning Curves
An exponential function may be applied to certain
types of learning processes with the model:
Q(t )  C  Ae
 kt
0  t  
C, A, k are positive constants
y
y=C
 0,C  A
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Suppose that the temperature T, in degrees Fahrenheit,
of an object after t minutes can be modeled using the
following equation:
0.3t
T (t )  200 150e
1. Find the temperature of the object after 5 minutes.
0.3(5)
T (5)  200 150e
 166.5
2. Find the time it takes for the temperature of the
object to reach 190°.
190  200 150e0.3t
1/15  e0.3t
t
ln 1/15 
0.3
 9 min.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Logistic Growth Model
An exponential function may be applied to a
logistic growth model:
A
Q(t ) 
 kt
1  Be
0  t  
A, B, k are positive constants
y
y=A
A 

0,


 1 B 
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
The number of people R, in a small school district
who have heard a particular rumor after t days can be
modeled by:
2400
R(t ) 
1  1199e kt
If 10 people know the rumor after 1 day, find the
number who heard it after 6 days.
2400
10 
1  1199e k (1)
1  1199ek  240
 239 
k   ln 
  1.613
 1199 
…
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
2400
R(t ) 
1.613t
1  1199e
So
2400
R(6) 
 2232 people
1.613(6)
1  1199e
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
The number of soldiers at Fort MacArthur who
contracted influenza after t days during a flu epidemic is
approximated by the exponential model:
5000
Q(t ) 
 kt
1  1249e
If 40 soldiers contracted the flu by day 7, find how
many soldiers contracted the flu by day 15.
5000
40 
1  1249e k (7)
1  1249e7k  125
1  124 
k   ln 
  0.33
7  1249 
…
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
5000
Q(t ) 
0.33t
1  1249e
So
5000
Q(15) 
 508 people
0.33(15)
1  1249e
So approximately 508 soldiers contracted
the flu by day 15.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.