Exponential functions 123
lne^3 x=ln7
i.e. 3 x=ln7
from which x=
1
3
ln7=0.6486,
correct to 4 decimal places.
Problem 10. 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 11. Evaluate the following: (a)
lne^2.^5
lg10^0.^5
(b)
5 e^2.^23 lg2. 23
ln2. 23
(correct to 3 decimal places)
(a)
lne^2.^5
lg10^0.^5
=
2. 5
0. 5
= 5
(b)
5 e^2.^23 lg2. 23
ln2. 23
=
5 ( 9. 29986607 ...)( 0. 34830486 ...)
( 0. 80200158 ...)
=20.194, correct to 3 decimal places.
Problem 12. Solve the equation 9= 4 e−^3 xto find
x, correct to 4 significant figures
Rearranging 9= 4 e−^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 13. 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 14. Solve the equation
2. 68 =ln
(
4. 87
x
)
to findx
From the definition of a logarithm, since
2. 68 =ln
(
4. 87
x
)
thene^2.^68 =
4. 87
x
Rearranging gives x=
4. 87
e^2.^68
= 4. 87 e−^2.^68
i.e. x=0.3339,
correct to 4 significant figures.
Problem 15. Solve
7
4
=e^3 xcorrect to 4
significant figures