Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

32 Higher Engineering Mathematics


From the last two examples we can conclude that:

logeex=x

This is useful whensolving equations involving expo-
nential functions. For example, to solve e^3 x=7, take
Napierian logarithms of both sides, which gives:
lne^3 x=ln7
i.e. 3 x=ln7

from which x=

1
3

ln7=0.6486,correct to 4
decimal places.

Problem 9. Evaluate the following, each correct
to 5 significant figures:

(a)

1
2

ln4.7291 (b)

ln7. 8693
7. 8693

(c)

3 .17ln24. 07
e−^0.^1762

.

(a)

1
2

ln4. 7291 =

1
2

( 1. 5537349 ...)=0.77687,
correct to 5 significant figures

(b)

ln7. 8693
7. 8693

=

2. 06296911 ...
7. 8693

=0.26215,

correct to 5 significant figures

(c)

3 .17ln24. 07
e−^0.^1762

=

3. 17 ( 3. 18096625 ...)
0. 83845027 ...
=12.027,
correct to 5 significant figures.

Problem 10. Evaluate the following:

(a)

lne^2.^5
lg10^0.^5

(b)

5e^2.^23 lg2. 23
ln2. 23

(correct to 3
decimal places).

(a)

lne^2.^5
lg10^0.^5

=

2. 5
0. 5

= 5

(b)

5e^2.^23 lg2. 23
ln2. 23

=

5 ( 9. 29986607 ...)( 0. 34830486 ...)
0. 80200158 ...
=20.194, correct to 3 decimal places.

Problem 11. Solve the equation: 9=4e−^3 xto
findx, correct to 4 significant figures.

Rearranging 9=4e−^3 xgives:
9
4

=e−^3 x

Taking the reciprocal of both sides gives:
4
9

=
1
e−^3 x

=e^3 x
Taking Napierian logarithms of both sides gives:

ln

(
4
9

)
=ln(e^3 x)

Since logeeα=α,thenln

(
4
9

)
= 3 x

Hence, x=

1
3

ln

(
4
9

)
=

1
3

(− 0. 81093 )=− 0. 2703 ,

correct to 4 significant figures.

Problem 12. Given 32= 70

(
1 −e−

t
2

)
determine
the value oft, correct to 3 significant figures.

Rearranging 32= 70 ( 1 −e−

t

(^2) )gives:
32
70
= 1 −e−
t
2
and e−
t
(^2) = 1 −
32
70


38
70
Taking the reciprocal of both sides gives:
e
t
(^2) =
70
38
Taking Napierian logarithms of both sides gives:
lne
t
(^2) =ln
(
70
38
)
i.e.
t
2
=ln
(
70
38
)
from which,t=2ln
(
70
38
)
= 1. 22 , correct to 3 signifi-
cant figures.
Problem 13. Solve the equation: 2. 68 =ln
(
4. 87
x
)
to findx.
From the definition of a logarithm, since
2. 68 =ln
(
4. 87
x
)
then e^2.^68 =
4. 87
x
Rearranging gives: x=
4. 87
e^2.^68
= 4 .87e−^2.^68
i.e. x= 0. 3339 , correct to 4
significant figures.
Problem 14. Solve
7
4
=e^3 xcorrect to 4 signi-
ficant figures.

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