Chapter 4 Exponential and Logarithmic
Functions (指数函数和对数函数)
In this Chapter, we will encounter
some important concepts
 Exponential Functions(指数函数)
 Logarithmic Functions(对数函数)
 Differentiation of Logarithmic and
Exponential Functions
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Section 4.1 Exponential Functions
Exponential function(指数函数): If b is a
positive number other than 1 (b>0, b≠1), there is
a unique function called the exponential function
with base b that is defined by
f(x)=bx for every real number x
NOTE: Such function can be used to describe exponential
and logistic growth and a variety of other important
quantities.
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Definition of b n for Rational Values of n (and b>0)
Integer powers: If n is a positive integer,
bn  b
b


b



n facters
Fractional powers: If n and m are positive integers,
b
where
m
n/m
  b
m
n
 b
m
n
b denotes the positive mth root.
Negative powers:
b
n
1
 n
b
0
b
1
Zero power:
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Example 1
Solution:
4
Figure below shows graphs of various members of the
x
family of exponential functions y  b
b
p
(
x
)

x
NOTE: Students often confuse the power function
x
f
(
x
)

b
with the exponential function
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6
7
Definition: The natural exponential function is
Where
n
10
100
2.59374
2.70481
1000
2.71692
10,000
2.711815
100,000
2.71827
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Continuous Compounding of Interest(连续利息)
If P is the initial investment (the principal) and r is the interest
rate (expressed as a decimal), the balance B after the interest is
added will be
B=P+Pr=P(1+r)
dollars
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10
11
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Example 2
Suppose $1,000 is invested at an annual interest rate of 6%.
Compute the balance after 10 years if the interest is compounded
a. Quarterly
b. Monthly
c. Daily
d. Continuously
Solution:
a. To compute the balance after 10 years if the interest is
compounded quarterly, using the formula
 r
B(t )  p1  
 k
kt
with t=10, p=1,000, r=0.06, and k=4:
40
 0.06 
B(10)  1,0001 
  $1,814.02
4


to be continued
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b. This time, take t=10, p=1,000, r=0.06, and k=12 to get
120
 0.06 
B(10)  1,0001 

12 

 $1,819.40
c. Take t=10, p=1,000, r=0.06, and k=365 to obtain
 0.06 
B(10)  1,0001 

365


3, 650
 $1,822.03
d. For continuously compounded interest use the formula B(t )  pert
with t=10, p=1,000, and r=0.06:
B(10)  1,000e0.6  $1,822.12
This value, $1,822.12, is an upper bound for the possible balance.
No matter how often interest is compounded, $1,000 invested at an
annual interest rate of 6% can not grow to more than $1,822.12 in
10 years.
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Present Value (现值) & Future Value
Present Value is the interest accumulation process in
reverse. Rather than adding interest to a principal to
determine a sum, it is in effect subtracted from a sum to
determine a principal.
The present value P of future value F with interest
rate r per conversion period for n period is given by
the formula
is the present value interest factor (PVIFn) (or discount
factor) of n periods.
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Present Value (现值) & Future Value
i = annualized interest rate, t = number of years
m = compound periods per year
If the interest rate is i*100% per year and the interest is
compounded continuously for t years, then the present
value P for a given amount of future value F can be
calculated from
where exp(−it) is the continuous PVIF.
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Present value of $1.00 when interest (i=5%) is
compounded quarterly for a year
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Example 3
Sue is about to enter college. When she graduates 4 years from
now, she wants to take a trip to Europe that she estimates will cost
$5,000. How much should she invest now at 7% to have enough
for the trip if interest is compounded:
a. Quarterly
b. Continuously
Solution:
The required future value is F=$5,000 in t=4 years with r=0.07.
a. If the compounding is quarterly, then k=4 and the present value
is
to be continued
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0.07 

P  5,0001 

4 

4 ( 4 )
 $3,788.08
b. For continuous compounding, the present value is
P  5,000e0.07 ( 4)  $3,778.92
Thus, Sue would have to invest about $9 more if interest is
compounded quarterly than if the compounding is continuous.
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Example4 The present value of a future expense.
Suppose that you plan to buy a luxury car four years from
now and that car will cost $55,000 at that time. If your bank’s
savings deposit interest rate is 5% per year, what is the
amount that you have to deposit now (present value) in your
savings account in order to grow enough interest and with
the principal to buy your dream car in the future?
Assume the interest is compounded: (a) annually;
(b) quarterly; (c) weekly; or (d) continuously, for a year.
(e) Show your results in a table of PV calculation with
separate FV, PVIF, and PV columns.
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Answer:
(a) annually P = $55,000(1 + 0.05)−4 = $45,248.64;
(b) quarterly P = $55,000(1 + 0.05/4)−4∗4 = $45,086.05;
(c) weekly P = $55,000(1 + 0.05/52)−52∗4 = $45,034.52;
(d) continuously P = $55, 000e−0.05∗4 = $45,030.19;
(e)
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Pricing the Cash Flows of a Bond
Application of present value
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23
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Coupon Bond valuation equation
Price of a bond is equal to the sum of present values of future coupons
plus that of the redemption value.
For a typical bond that promises to pay a fixed
coupon payment of C/2 every six months and
repay the par amount FV (face value) at maturity,
current price P is:
where r /2 is the appropriate semiannual yield rate and n is the remaining
life of the bond measured in the unit of the coupon payment period (six months).
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v=1/(1+i)
Present value of annuity
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Example: How to value a coupon
bond?
1
Suppose a bond has a face value of $1000, will
pay semi-annual interest at the (coupon) rate of
5%, and will be due 10 years after issue.
This means that the owner of one such bond will
receive $25 (=1000*5%/2) every six months for a
total of 20 payments, together with a payment of
$1000 10 years after the date of issue.
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Example: How to value a coupon
bond?
2
If the purchaser wishes to receive a 6% return,
compounded semiannually, then the present value at
issue is
A purchaser content with a return of 4% compounded
semi-annually would value the bond at
Price of a bond is equal to the present value of future coupons
plus that of the redemption value.
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Sensitivity of Bond Prices to
Interest Rate Movements
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idV
Bond price elasticity =
Vdi
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Discounted Cash-flow Valuation:
A share of preferred stock is like a perpetual bond, and
has no maturity. It receives a pre-specified regular dividend.
Present value of preferred stock =
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指数增长和指数下降
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Section 4.2 Logarithmic Functions(对数函数)
34
35
Example 4
Use logarithm rules to rewrite each of the following expressions
in terms of log5 2 and log
. 53
5
log
a.
b. log5 8
c. log5 36
5 
3
Solution:
a.
5
log5    log5 5  log5 3
 3
 1  log5 3
quot ientrule
since log5 5  1
b. log5 8  log5 23  3 log5 2
c.
power rule
log5 36  log5 (2 232 )  log5 2 2  log5 32
 2log5 2  2 log5 3
product rule
power rule
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Example 5
Solution:
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Exponential & Logarithmic functions:
important limits

Lim (x0) (exp(x) – 1 ) / x = 1



Or exp(x) 1 + x as x0 , and
Lim (x0) ln(1+x) / x = 1
From these, we can prove


d/dx ( exp(x) ) = exp(x) ,
d/dx ( ln(x) ) = 1/x
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Section 4.3 Differentiation of Logarithmic
and Exponential Function
Example 6
Solution:
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Example 7
42
Solution:
43
Differentiate both sides of the equation
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Example 8
Solution:
45
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Example 9
Solution:
47
Taking the derivatives of some complicated functions can
be simplified by using logarithms. This is called
logarithmic differentiation.
Example 10
Solution:
to be continued
48
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Example 11
Differentiate each of these function
a.
y  2x
b.
y  log3 x
Solution:
a. To differentiate
follows:
y  2 x , we use logarithmic differentiation as
to be continued
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b.
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52
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The relative rate of change of a quantity Q(x) can be
computed by finding the derivative of lnQ.
d
Q '( x )
(ln Q) 
dx
Q( x)
54
Example 12
Solution:
to be continued
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Summary
 Exponential Functions, Basic Properties of Exponential
Functions, The Natural Exponential Base e.
 Compound Interest, Continuously Compounded Interest,
Present Value.
 Exponential Growth and Decay.
 Logarithmic Functions, The Natural Logarithm.
 Differentiation of logarithmic and Exponential
Functions.
 Optimal holding time.
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Trigonometric functions: a very
important limit
From this we can prove:
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Derivatives of trigonometric functions

Using the previous results we can derive




d/dx sin(x) = cos(x)
d/dx cos(x) = - sin(x)
d/dx tan(x) = ?
d/dx sec(x) = ?
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Hyperbolic functions




sinh (x) = ( exp(x) – exp(-x) )/ 2
cosh (x) = ( exp(x) + exp(-x) )/ 2
d/dx sinh(x) = cosh(x)
d/dx cosh(x) = sinh(x)
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Indeterminate forms

Expressions of the form 0/0, ∞/∞, 0 ×∞,
∞−∞, 0^∞ and ∞^0 are called indeterminate
forms.
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L’Hospital’s Rule – for indeterminate
forms
If f’(x) and g’(x) exist, and if g’(x) is nonzero,
then we have the following result, known as
L’Hospital’s rule:
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Examples:
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