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.