Higher Engineering Mathematics

(Greg DeLong) #1
34 NUMBER AND ALGEBRA

(a)

ln e^2.^5
lg 10^0.^5

=

2. 5
0. 5

= 5

(b)


4e^2.^23 lg 2. 23
ln 2. 23

=

4(9. 29986607 ...)(0. 34830486 ...)
0. 80200158 ...
=16.156, correct to 3 decimal places

Problem 21. Solve the equation 7=4e−^3 xto
findx, correct to 4 significant figures.

Rearranging 7=4e−^3 xgives:
7
4

=e−^3 x

Taking the reciprocal of both sides gives:
4
7

=

1
e−^3 x

=e^3 x

Taking Napierian logarithms of both sides gives:

ln

(
4
7

)
=ln (e^3 x)

Since logeeα=α, then ln

(
4
7

)
= 3 x.

Hence

x=

1
3

ln

(
4
7

)
=

1
3

(− 0 .55962)

=−0.1865, correct to 4 significant figures

Problem 22. Given 20=60(1−e

−t

(^2) ) deter-
mine the value of t, correct to 3 significant
figures.
Rearranging 20=60(1−e
−t
(^2) ) gives:
20
60
= 1 −e
−t
2
and
e
−t
(^2) = 1 −
20
60


2
3
Taking the reciprocal of both sides gives:
e
t
(^2) =
3
2
Taking Napierian logarithms of both sides gives:
ln e
t
(^2) =ln
3
2
i.e.
t
2
=ln
3
2
from which,t=2ln
3
2
=0.811, correct to 3 signifi-
cant figures
Problem 23. Solve the equation
3. 72 =ln
(
5. 14
x
)
to findx.
From the definition of a logarithm, since
3. 72 =ln
(
5. 14
x
)
then e^3.^72 =
5. 14
x
Rearranging gives:
x=
5. 14
e^3.^72
= 5 .14 e−^3.^72
i.e.x=0.1246, correct to 4 significant figures
Now try the following exercise.
Exercise 21 Further problems on evaluat-
ing Napierian logarithms



  1. Evaluate, correct to 4 decimal places


(a) ln 1.73 (b) ln 541.3 (c) ln 0. 09412
[(a) 0.5481 (b) 6.2940 (c)−2.3632]


  1. Evaluate, correct to 5 significant figures.


(a)

2 .946 ln e^1.^76
lg 10^1.^41

(b)

5e−^0.^1629
2ln0. 00165

(c)

ln 4. 8629 −ln 2. 4711
5. 173
[(a) 3.6773 (b)−0.33154 (c) 0.13087]
In Problems 3 to 7 solve the given equations,
each correct to 4 significant figures.




    1. 5 =4e^2 t [−0.4904]





    1. 83 = 2 .91 e−^1.^7 x [−0.5822]



  1. 16=24(1−e


−t

(^2) ) [2.197]





    1. 17 =ln




( x

4. 64

)
[816.2]


  1. 3.72 ln


(
1. 59
x

)
= 2. 43 [0.8274]
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