Chapter 4 –Solution
PROBLEMS
1. Find all real numbers x that satisfy the equation 2 3 x  4 x .
Solution (e.g. 4.1.3)
The equation 2 3 x  4 x is satisfied if and only if
23 x  2 2 x
3  x  2x
x 1
Thus, 2 3 x  4 x if only if x  1 .
2. POPULATION GROWTH it is estimated that the population of a certain country grows
exponentially. If the population was 60 million in 1997 and 90 million in 2002, what will the
population be in 2012?
Solution (e.g. 4.1.7)
Let P (x ) denote the population measured in million unit’s t years after the base year 1997. Then P
is a function of the form
P( x)  60e kt
Since the population was 90 million in 2002, it follows that
3
90  60e 5 k
e5k 
Or
2
In 2012, the population will be
3
P(15)  60e15k  60(e 5 k ) 3  60( ) 3  202.5 million.
2
4


x
3. Use logarithmic rules to simplify ln 

3
2
 x 1 x 
Solution (e.g. 4.2.6)
 4 x
ln 
3
2
 x 1 x



x1 / 4

ln

 x 3 (1  x 2 )1/ 2 



1
1
 ln x  (3 ln x  ln( 1  x 2 ))
4
2
1
1
1
 ln x  3 ln x  ln( 1  x)  ln( 1  x)
4
2
2
1
4. Solve the equation ln x  (ln 16  2 ln 2) for x.
3
Solution (e.g. 4.2.8)
Using the logarithm rules we find that
1
1
(ln 16  2 ln 2)  (4 ln 2  2 ln 2)  2 ln 2  ln 4
3
3
That is , ln x  ln 4 . Take the natural logarithm of each side of this equation to get
e ln x  e ln 4 or x  4
1
Math-Chapter 4
5. COMPOUND INTEREST How quickly will money double if it is invested at an annual interest
rate of 6% compounded continuously?
Solution (e.g. 4.2.10)
Apply the doubling time theory (page 311) to find that the money will be double when
ln 2 ln 2
t

 11.55 years.
k
0.06
6. ARCHAEOLOGY Tests of an artifact discovered at the Debert site in Nova Scotia show that 28%
of the original 14C is still present. Approximately how old is the artifact?
Solution (e.g. 4.2.12)
The age of the artifact is the value of t for which R(t )  28%R0 , that is for which
0.28 R0  R0 e  kt
ln 0.28
t
So
k
ln 2
We know half-life h satisfies h 
, and since h  5730 years, we have
k
ln 0.28
ln 0.28
ln 0.28
t


 5730  10523
ln 2
k
ln 2
h
So the artifact is approximately 10,523 years old.
7. Differentiate the function f ( x)  ( x 2  3x  5)e 6 x
Solution (e.g. 4.3.8, 4.3.9)
f ' ( x)  (2 x  3)e6 x  ( x 2  3x  5)e6 x  6
 (6 x 2  20 x  33)e6 x
8. Find the largest and smallest values of the function f ( x) 
ln( x  1)
over interval [0, 2]
x 1
Solution (e.g. 4.3.9)
f ' ( x) 
ln( x  1)  ( x  1)
( x  1) 2
1
x  1  ln( x  1)  1
( x  1) 2
so f ' ( x)  0 when ln( x  1)  1 , that is x 1  e , or x  e  1 [0,2] . Evaluating f (x) at the critical
number x  e  1 and at the endpoints of the interval, x  0 and x  2 , we find that
1
ln 3
f (0)  0, f (e  1)   0.368, f (2) 
 0.366
e
3
Thus, f (x) has its largest value at x  e  1, and has its smallest value at x  0 .
9. Given f ( x)  5 x , use the logarithmic differentiation to find the derivative f ' ( x) .
Solution (e.g. 4.3.12, 4.3.13)
Use logarithm differentiation as follows
2
f ( x)  5 x
ln f ( x)  x 2 ln 5
2
2
Math-Chapter 4
f ' ( x)
 2 ln 5  x
f ( x)
f ' ( x)  2 ln 5  x  f ( x)
f ' ( x)  2 ln 5  x  5x
2
10. The cost of producing x units of a particular commodity is C ( x)  x 2  10 xe x .
(a) Find the marginal cost C ' ( x)
(b) Determine the level of production x for which the average cost A( x)  C ( x) / x is minimized.
Solution (e.g. 4.3.6)
a. The marginal cost C ' ( x) is
C ' ( x)  2 x  10e  x  10 xe x  (1)  2 x  10e  x  10 xe x
b. The average cost A( x)  C ( x) / x is
C ( x)
A( x) 
 x  10e  x
x
Its derivative is A' ( x)  1  10e  x which is zero when 1  10e  x  0 , that is x  ln 10 . Since
A' ( x)  0 for 0  x  ln 10 , and A' ( x)  0 for ln 10  0 , it follows that
A(ln 10)  ln 10  10e  ln10  ln 10  1
is minimum of A(x ) .
11. THE SPREAD OF AN EPIDEMIC
Public health records indicate that t weeks after the
2
outbreak of a certain form of influenza, approximately f (t ) 
thousand people had
1  3e 0.8t
caught the disease.
b. How many people had the disease initially?
c. How many had caught the disease by the end of 3 weeks?
d. If the trend continues, approximately how many people in all will contract the disease?
Solution (e.g. 4.4.5)
b. Since f (0)  0.5 , there were 500 people had caught the disease
initially.
Y=2
c.
f (3) 
2
 1.572
1  3e 2.4
Thus , there were 1572 people had caught the disease by the end of
3 weeks.
2
ln3/0.8
2
d. Since f (t ) approaches 2 as t increases without bound, it follows
that approximately 2000 people will contract the disease.
12. EFFICIENCY The daily output of a worker who has been on
the job for t weeks is given by a function of the form Q(t )  40  Ae  kt . Initially the worker could
ln3/0.8
produce 20 units a day, and after 1 week the worker can produce 30 units a day. How many units
will the worker produce per day after 3 weeks?
3
Math-Chapter 4
Solution (e.g. 4.4.4)
Since the worker could produce 20 units a day initially and 30 units a day after 1 week, that is,
Q(0)  20 and Q(1)  30 , we can get the equations
40  A  20 and 40  Ae k  30
Then solve A and k from the equations, A  20 and k  ln 2 . That is, Q(t )  40  20e  ln 2t . After 3
weeks , the worker will produce
20
Q(3)  40  20e ln 23  40  3  37.5 units per week.
2
13. OPTIMAL HOLDING TIME Suppose your family owns a rare book whose value t years from
now will be V (t )  200e 2t dollars. If the prevailing interest rate remains constant at 6% per year
compounded continuously, when will it be most advantageous for your family to sell the book and
invest the proceeds?
Solution (e.g. 4.4.3)
In t years, the marker price of the book will be V (t )  200e 2t . The present value of this book is
P(t )  200e 2t 0.06t dollars.
1
P' (t )  200e 2t 0.06t (
 0.06)
2t
Thus, P ' (t ) is undefined when t  0 and P' (t )  0 when
2
1
 1 
 0.06  0 or t  
  138.89
2t
 0.06 2 
Since P ' (t ) >0 for and P ' (t ) <0 for t  138.89 , it follows that the present value is maximized in
approximately 138.89 years.
14. OPTIMAL AGE FOR REPRODUCTION Suppose that for a particular semelparous organism,
the likelihood of an individual surviving to age x years is p( x)  e 0.2 x and that the number of
female births to an individual at age x is f ( x)  5x 0.9 . What is the ideal age for reproduction for an
individual organism of this species? (See Example 4.4.6)
Solution (e.g. 4.4.6)
The per capita rate of increase function for this model is
ln[ e 0.2 x (5 x 0.9 )]
R( x) 
 x 1 (ln e 0.2 x  ln 5  ln x 0.9 x )
x
 0.2  (ln 5  0.9 ln x) x 1
0.9 1  ln 5  0.9 ln x  0.9
R' ( x)  0  (ln 5  0.9 ln x)(  x 2 ) 
x 
So
x
x2
0.9  ln 5
Therefore R' ( x)  0 when  ln 5  0.9 ln x  0.9  0 , that is ln x 
, or
0 .9
xe
4
0.9ln 5
0.9
 0.4546 .
Math-Chapter 4
1.8 ln x  2 ln 5  2.7
x3
Since R' ' (0.4546)  0 , it follows that the largest value of R (x ) occurs when x  0.4546 . Thus , the
optimal age for an individual organism to reproduction is when it is 0.4546 years old.
R' ' ( x) 
5