Exponential functions 123
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 10. 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 ln24.07
e −0.1762 =
3.17(3.18096625 ...)
0.83845027 ...
= 12.027,
correct to 5 significant figures.
Problem 11. 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 12. 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 13. 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 14. 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 15. Solve
7
4
= e 3x correct to 4
significant figures