solutions to exercises, section 4

Instructor’s Solutions Manual, Section 4.4 Exercise 1

Solutions to Exercises, Section 4.4

For Exercises 1–14, estimate the indicated value without using acalculator.

1. ln 1.003

solutionln 1.003 = ln(1+ 0.003) ≈ 0.003


2. ln 1.0007

solutionln 1.0007 = ln(1+ 0.0007) ≈ 0.0007


3. ln 0.993

solutionln 0.993 = ln

(1+ (−0.007)

) ≈ −0.007


4. ln 0.9996

solutionln 0.9996 = ln

(1+ (−0.0004)

) ≈ −0.0004


5. ln 3.0012− ln 3

solution

ln 3.0012− ln 3 = ln3.0012

3= ln 1.0004

= ln(1+ 0.0004)

≈ 0.0004


6. ln 4.001− ln 4

solution

ln 4.001− ln 4 = ln4.001

4= ln 1.00025

= ln(1+ 0.00025)

≈ 0.00025


7. e0.0013

solutione0.0013 ≈ 1+ 0.0013 = 1.0013


8. e0.00092

solutione0.00092 ≈ 1+ 0.00092 = 1.00092


9. e−0.0083

solutione−0.0083 ≈ 1+ (−0.0083) = 0.9917


10. e−0.00046

solutione−0.00046 ≈ 1+ (−0.00046) = 0.99954


11.e9

e8.997

solution

e9

e8.997= e9−8.997 = e0.003 ≈ 1+ 0.003 = 1.003


12.e5

e4.984

solution

e5

e4.984= e5−4.984 = e0.016 ≈ 1+ 0.016 = 1.016


13.(e7.001

e7

)2

solution

(e7.001

e7

)2 = (e7.001−7)2 = (e0.001)2

= e0.002

≈ 1+ 0.002 = 1.002


14.(e8.0002

e8

)3

solution

(e8.0002

e8

)3 = (e8.0002−8)3 = (e0.0002)3

= e0.0006

≈ 1+ 0.0006 = 1.0006


15. Estimate the value of

(1+ 3

10100

)(10100).

[Your calculator will be unable to evaluate directly the expressions in thisexercise and the next five exercises. Thus you will need to do more thanbutton pushing for these exercises.]

solution(1+ 3

10100

)(10100) ≈ e3 ≈ 20.09



(1+ 5

1090

)(1090).

solution(1+ 5

1090

)(1090) ≈ e5 ≈ 148.4


17. Estimate the value of (1− 4

980

)(980).

solution(1− 4

980

)(980) ≈ e−4 ≈ 0.01832


18. Estimate the value of (1− 2

899

)(899).

solution(1− 2

899

)(899) ≈ e−2 ≈ 0.1353



(1+ 10−1000)2·101000.

solution

(1+ 10−1000)2·101000 =((1+ 10−1000)101000

)2

=((

1+ 1

101000

)101000)2

≈ e2

≈ 7.389



(1+ 10−100)3·10100.

solution

(1+ 10−100)3·10100 =((1+ 10−100)10100

)3

=((

1+ 1

10100

)10100)3

≈ e3

≈ 20.0855


21. Estimate the slope of the line containing the points

(5, ln 5) and(5+ 10−100, ln(5+ 10−100)

).

solution The slope of the line containing the points

(5, ln 5) and(5+ 10−100, ln(5+ 10−100)

)is obtained in the usual way by taking the ratio of the difference of thesecond coordinates with the difference of the first coordinates:

ln(5+ 10−100)− ln 5

5+ 10−100 − 5= ln

(1+ 1

5 · 10−100)10−100

≈15 · 10−100

10−100

= 15 .

Thus the slope of the line in question is approximately 15 .


22. Estimate the slope of the line containing the points

(4, ln 4) and(4+ 10−1000, ln(4+ 10−1000)

).

solution The slope of the line containing the points

(4, ln 4) and(4+ 10−1000, ln(4+ 10−1000)

)is obtained in the usual way by taking the ratio of the difference of thesecond coordinates with the difference of the first coordinates:

ln(4+ 10−1000)− ln 4

4+ 10−1000 − 4= ln

(1+ 1

4 · 10−1000)10−1000

≈14 · 10−1000

10−1000

= 14 .

Thus the slope of the line in question is approximately 14 .


23. Suppose t is a small positive number. Estimate the slope of the linecontaining the points (4, e4) and (4+ t, e4+t).

solution The slope of the line containing (4, e4) and (4+ t, e4+t) isobtained in the usual way by taking the ratio of the difference of thesecond coordinates with the difference of the first coordinates:

e4+t − e4

4+ t − 4= e4(et − 1)

t

≈ e4(1+ t − 1)t

= e4

≈ 54.598

Thus the slope of the line in question is approximately 54.598.


24. Suppose r is a small positive number. Estimate the slope of the linecontaining the points (7, e7) and (7+ r , e7+r ).

solution The slope of the line containing (7, e7) and (7+ r , e7+r ) isobtained in the usual way by taking the ratio of the difference of thesecond coordinates with the difference of the first coordinates:

e7+r − e7

7+ r − 7= e7(er − 1)

r

≈ e7(1+ r − 1)r

= e7

≈ 1096.63



25. Suppose r is a small positive number. Estimate the slope of the linecontaining the points (e2,6) and (e2+r ,6+ r).solution The slope of the line containing (e2,6) and (e2+r ,6+ r) isobtained in the usual way by taking the ratio of the difference of thesecond coordinates with the difference of the first coordinates:

6+ r − 6e2+r − e2

= re2(er − 1)

≈ re2(1+ r − 1)

= 1e2

≈ 0.135



26. Suppose b is a small positive number. Estimate the slope of the linecontaining the points (e3,5+ b) and (e3+b,5).

solution The slope of the line containing (e3,5+ b) and (e3+b,5) isobtained in the usual way by taking the ratio of the difference of thesecond coordinates with the difference of the first coordinates:

5+ b − 5e3 − e3+b =

be3(1− eb)

≈ be3(1− (1+ b))

= − 1e3

≈ −0.0498

Thus the slope of the line in question is approximately −0.0498.


27. Find a number r such that

(1+ r

1090

)(1090) ≈ 5.

solution If r is not a huge number, then

(1+ r

1090

)(1090) ≈ er .

Thus we need to find a number r such that er ≈ 5. This implies thatr ≈ ln 5 ≈ 1.60944.


28. Find a number r such that

(1+ r

1075

)(1075) ≈ 4.

solution If r is not a huge number, then

(1+ r

1075

)(1075) ≈ er .

Thus we need to find a number r such that er ≈ 4. This implies thatr ≈ ln 4 ≈ 1.38629.


29. Find the number c such that

area( 1x ,2, c) = 3.

solution We have

3 = area( 1x ,2, c)

= area( 1x ,1, c)− area( 1

x ,1,2)

= ln c − ln 2

= ln c2 .

Thus c2 = e3, which implies that c = 2e3 ≈ 40.171.


30. Find the number c such that

area( 1x ,5, c) = 4.

solution We have

4 = area( 1x ,5, c)

= area( 1x ,1, c)− area( 1

x ,1,5)

= ln c − ln 5

= ln c5 .

Thus c5 = e4, which implies that c = 5e4 ≈ 272.991.

Instructor’s Solutions Manual, Section 4.4 Problem 31

Solutions to Problems, Section 4.4

31. Show that1

1020 + 1< ln(1+ 10−20) <

11020

.

solution Using the inequality h1+h < ln(1+ h) < h with h = 10−20, we

have10−20

1+ 10−20< ln(1+ 10−20) < 10−20 = 1

1020.

Multiplying numerator and denominator of the term on the left by 1020,we can rewrite this as

11020 + 1

< ln(1+ 10−20) <1

1020.


32. (a) Using a calculator, verify that

log(1+ t) ≈ 0.434294t

for some small numbers t (for example, try t = 0.001 and thensmaller values of t).

(b) Explain why the approximation above follows from theapproximation ln(1+ t) ≈ t.

solution

(a) Using a calculator, we find the following values for log( 1+ t):

t log(1+ t) 0.434294t

10−3 4.34077× 10−4 4.34294× 10−4

10−4 4.34273× 10−5 4.34294× 10−5

10−5 4.34292× 10−6 4.34294× 10−6

10−6 4.34294× 10−7 4.34294× 10−7

This data indicates that

log(1+ t) ≈ 0.434294t

for small numbers t.

(b) If t is a small number, then


log(1+ t) = (log e)(loge(1+ t)

)≈ 0.434294 ln(1+ t)≈ 0.434294t.


33. (a) Using a calculator or computer, verify that

2t − 1 ≈ 0.693147t

for some small numbers t (for example, try t = 0.001 and thensmaller values of t).

(b) Explain why 2t = et ln 2 for every number t.

(c) Explain why the approximation in part (a) follows from theapproximation et ≈ 1+ t.

solution

(a) Using a computer, we find the following values for 2t :

t 2t − 1 0.693147t

10−3 6.93387× 10−4 6.93147× 10−4

10−4 6.93171× 10−5 6.93147× 10−5

10−5 6.93150× 10−6 6.93147× 10−6

10−6 6.93147× 10−7 6.93147× 10−7

This data indicates that

2t − 1 ≈ 0.693147t

for small numbers t.


(b)2t = eln 2t = et ln 2

(c) If t is a small number, then

2t − 1 = et ln 2 − 1

≈ (1+ t ln 2)− 1

= t ln 2

≈ 0.693147t.


34. Suppose x is a positive number.

(a) Explain why xt = et lnx for every number t.

(b) Explain whyxt − 1t

≈ lnx

if t is close to 0.

[Part (b) of this problem gives another illustration of why the naturallogarithm deserves the title “natural”.]

solution

(a)xt = elnxt = et lnx

(b) If t is a small number, then

xt − 1t

= et lnx − 1t

≈ (1+ t lnx)− 1t

= lnx.


35. (a) Using a calculator or computer, verify that

(1+ ln 10x )

x ≈ 10

for large values of x (for example, try x = 1000 and then largervalues of x).

(b) Explain why the approximation above follows from theapproximation (1+ r

x )x ≈ er .

solution

(a) Using a computer, we find the following values for (1+ ln 10x )

x:

x (1+ ln 10x )

x

103 9.97357

104 9.99735

105 9.99973

106 9.99997

This data indicates that(1+ ln 10

x )x ≈ 10

for large numbers x.

(b) If x is a large number, then

(1+ ln 10x )

x ≈ eln 10 = 10.


36. Using a calculator, discover a formula for a good approximation for

ln(2+ t)− ln 2

for small values of t (for example, try t = 0.04, t = 0.02, t = 0.01, andthen smaller values of t). Then explain why your formula is indeed agood approximation.

solution A calculator gives the following values for ln(2+ t)− ln 2:

t ln(2+ t)− ln 2

0.04 0.0198

0.02 0.00995

0.01 0.00499

0.001 0.0004999

0.0001 0.000049999

The table above suggests that

ln(2+ t)− ln 2 ≈ t2

for small values of t.

To see that this approximation is indeed valid, suppose t is a smallnumber. Then


ln(2+ t)− ln 2 = ln 2+t2

ln(1+ t2)

≈ t2 .


37. Show that for every positive number c, we have

ln(c + t)− ln c ≈ tc


solution Suppose c is a positive number and t is a very smallnumber. Then

ln(c + t)− ln c = ln c+tc

ln(1+ tc )

≈ tc .


38. Show that for every number c, we have

ec+t − ec ≈ tec


solution Suppose c is a number and t is a very small number. Then

ec+t − ec = etec − ec

= (et − 1)ec

≈ tec.


39. Show that if t > 0, then 1+ t < et .[This problem and the next problem combine to show that

1+ t < et < (1+ t)1+t

if t > 0.]

solution Suppose t > 0. Then

ln(1+ t) < t.

Thuseln(1+t) < et,

which can be rewritten as1+ t < et.


40. Show that if t > 0, then et < (1+ t)1+t .solution Suppose t > 0. Then

t1+ t < ln(1+ t).

Thust < (1+ t) ln(1+ t) = ln(1+ t)1+t,

which implies thatet < eln(1+t)1+t ,

which can be rewritten as

et < (1+ t)1+t.


41. Show that if x > 0, then(1+ 1

x)x < e.

[This problem and the next problem combine to show that

(1+ 1

x)x < e < (1+ 1

x)x+1

if x > 0.]

solution Suppose x > 0. Take h = 1x in the result of Problem 39,

getting1+ 1

x < e1/x.

Now raise both sides of the inequality above to the power x, getting

(1+ 1

x)x < e.


42. Show that if x > 0, then e <(1+ 1

x)x+1

.

solution Suppose x > 0. Take h = 1x in the result of Problem 40,

getting

e1/x <(1+ 1

x)1+ 1

x .

Now raise both sides of the inequality above to the power x, getting

e <(1+ 1

x)x+1.


43. (a) Show that1.01100 < e < 1.01101.

(b) Explain why1.01100 + 1.01101

2

is a reasonable estimate of e.

solution

(a) Take x = 100 in the results of the two previous problems to get thedesired inequalities.

(b) Part (a) states that e is between 1.01100 and 1.01101. Thus we averagethese two numbers to obtain an estimate for e.A computer shows that

1.01100 + 1.01101

2≈ 2.71834

ande ≈ 2.71828.

Both of these approximations round off to 2.7183, providing agreementto four digits after the decimal point.


44. Show thatarea( 1

x ,1b ,1) = area( 1

x ,1, b)

for every number b > 1.

solution Suppose b > 1. Define a function f on the interval [ 1b ,1] by

f(x) = 1bx ;

then define a function h on the interval [1, b] by

h(x) = f (xb ) = 1

bxb= 1

x .

Our results on stretching a graph horizontally (Section 1.3) show thatthe graph of h is obtained by horizontally stretching the graph of f bya factor of b. The Area Stretch Theorem (Section 4.2) now implies thatthe area under the graph of h, above the x-axis, and between the linesx = 1 and x = b equals b times the area under the graph of f , abovethe x-axis, and between the lines x = 1

b and x = 1. In other words,

(44.a) area( 1x ,1, b) = b area( 1

bx ,1b ,1).

Now define a function g on the interval [ 1b ,1] by

g(x) = bf(x) = b · 1bx = 1

x .

Our results on stretching a graph vertically (Section 1.3) show that thegraph of g is obtained by vertically stretching the graph of f by a factor


of b. The Area Stretch Theorem (Section 4.2) now implies that the areaunder the graph of g, above the x-axis, and between the lines x = 1

band x = 1 equals b times the area under the graph of f , above thex-axis, and between the lines x = 1

b and x = 1. In other words,

(44.b) area( 1x ,

1b ,1) = b area( 1

bx ,1b ,1).

Equations (44.a) and (44.b) have the same right sides, and thus their leftsides are equal. Hence

area( 1x ,

1b ,1) = area( 1

x ,1, b).


45. Show that if 0 < a < 1, then

area( 1x , a,1) = − lna.

solution Suppose 0 < a < 1. Then

area( 1x , a,1) = area( 1

x ,1,1a) (by taking b = 1

a in the previous problem)

= ln 1a

= − lna.


46. Show thatarea( 1

x , a, b) = area( 1x ,1,

ba)

whenever 0 < a < b.

solution Suppose 0 < a < b. Define a function f on the interval[a, b] by

f(x) = ax ;

then define a function h on the interval [1, ba] by

h(x) = f(ax) = 1x .

Our results on stretching a graph horizontally (Section 1.3) show thatthe graph of h is obtained by horizontally stretching the graph of f bya factor of 1

a . The Area Stretch Theorem (Section 4.2) now implies thatthe area under the graph of h, above the x-axis, and between the linesx = 1 and x = b

a equals 1a times the area under the graph of f , above

the x-axis, and between the lines x = a and x = b. In other words,

(46.a) area( 1x ,1,

ba) =

1a

area( ax , a, b).

Now define a function g on the interval [a, b] by

g(x) = 1af(x) = 1

a · ax = 1x .

Our results on stretching a graph vertically (Section 1.3) show that thegraph of g is obtained by vertically stretching the graph of f by a factor


of 1a . The Area Stretch Theorem (Section 4.2) now implies that the area

under the graph of g, above the x-axis, and between the lines x = aand x = b equals 1

a times the area under the graph of f , above thex-axis, and between the lines x = a and x = b. In other words,

(46.b) area( 1x , a, b) = 1

a area( ax , a, b).

Equations (46.a) and (46.b) have the same right sides, and thus their leftsides are equal. Hence

area( 1x , a, b) = area( 1

x ,1,ba).


47. Show that

area( 1x , a, b) = ln

ba

whenever 0 < a < b.

solution

area( 1x , a, b) = area( 1

x ,1,ba) (from the previous problem)

= ln ba


48. Show that sinhx ≈ x if x is close to 0.[The definition of sinh was given before Exercise 52 in Section 4.3.]

solution Suppose x is close to 0. Then

sinhx = ex − e−x2

≈ (1+ x)− (1− x)2

= x.

solutions to exercises, section 4

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