32 Higher Engineering Mathematics
From the last two examples we can conclude that:
log e e
x
= x
This is useful when solving equations involving exponential functions. For example, to solve e 3x = 7, take
Napierian logarithms of both sides, which gives:
ln e 3x = ln 7
i.e.
3x = ln 7
from which
x =
1
3
ln 7 = 0.6486, correct to 4
decimal places.
Problem 9. Evaluate the following, each correct
to 5 significant figures:
(a)
1
2
ln 4.7291 (b)
ln 7.8693
7.8693
(c)
3.17 ln 24.07
e −0.1762 .
(a)
1
2
ln 4.7291 =
1
2
(1.5537349 ...) = 0.77687,
correct to 5 significant figures
(b)
ln 7.8693
7.8693
=
2.06296911 ...
7.8693
= 0.26215,
correct to 5 significant figures
(c)
3.17 ln 24.07
e −0.1762 =
3.17(3.18096625 ...)
0.83845027 ...
= 12.027,
correct to 5 significant figures.
Problem 10. Evaluate the following:
(a)
ln e 2.5
lg 10 0.5 (b)
5e 2.23 lg 2.23
ln 2.23
(correct to 3
decimal places).
(a)
ln e 2.5
lg 10 0.5 =
2.5
0.5
= 5
(b)
5e 2.23 lg 2.23
ln 2.23
=
5(9.29986607 ...)(0.34830486 ...)
0.80200158 ...
= 20.194, correct to 3 decimal places.
Problem 11. Solve the equation: 9 = 4e −3x to
find x, correct to 4 significant figures.
Rearranging 9 = 4e −3x gives:
9
4
= e −3x
Taking the reciprocal of both sides gives:
4
9
=
1
e −3x = e 3x
Taking Napierian logarithms of both sides gives:
ln
4
9
= ln(e 3x )
Since log e e α = α, then ln
4
9
= 3x
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 of t , 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:
ln e
t
2 = ln
70
38
i.e.
t
2
= ln
70
38
from which, t = 2 ln
70
38
= 1.22, correct to 3 significant figures.
Problem 13. Solve the equation: 2.68 = ln
4.87
x
to find x.
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 3x correct to 4 significant figures.
From the last two examples we can conclude that:
log e e
x
= x
This is useful when solving equations involving exponential functions. For example, to solve e 3x = 7, take
Napierian logarithms of both sides, which gives:
ln e 3x = ln 7
i.e.
3x = ln 7
from which
x =
1
3
ln 7 = 0.6486, correct to 4
decimal places.
Problem 9. Evaluate the following, each correct
to 5 significant figures:
(a)
1
2
ln 4.7291 (b)
ln 7.8693
7.8693
(c)
3.17 ln 24.07
e −0.1762 .
(a)
1
2
ln 4.7291 =
1
2
(1.5537349 ...) = 0.77687,
correct to 5 significant figures
(b)
ln 7.8693
7.8693
=
2.06296911 ...
7.8693
= 0.26215,
correct to 5 significant figures
(c)
3.17 ln 24.07
e −0.1762 =
3.17(3.18096625 ...)
0.83845027 ...
= 12.027,
correct to 5 significant figures.
Problem 10. Evaluate the following:
(a)
ln e 2.5
lg 10 0.5 (b)
5e 2.23 lg 2.23
ln 2.23
(correct to 3
decimal places).
(a)
ln e 2.5
lg 10 0.5 =
2.5
0.5
= 5
(b)
5e 2.23 lg 2.23
ln 2.23
=
5(9.29986607 ...)(0.34830486 ...)
0.80200158 ...
= 20.194, correct to 3 decimal places.
Problem 11. Solve the equation: 9 = 4e −3x to
find x, correct to 4 significant figures.
Rearranging 9 = 4e −3x gives:
9
4
= e −3x
Taking the reciprocal of both sides gives:
4
9
=
1
e −3x = e 3x
Taking Napierian logarithms of both sides gives:
ln
4
9
= ln(e 3x )
Since log e e α = α, then ln
4
9
= 3x
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 of t , 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:
ln e
t
2 = ln
70
38
i.e.
t
2
= ln
70
38
from which, t = 2 ln
70
38
= 1.22, correct to 3 significant figures.
Problem 13. Solve the equation: 2.68 = ln
4.87
x
to find x.
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 3x correct to 4 significant figures.