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.