Basic Engineering Mathematics, Fifth Edition

124 Basic Engineering Mathematics

Taking natural logs of both sides gives

ln

7

4

=lne^3 x

ln

7

4

= 3 xlne

Sincelne=^1 , ln

7

4

= 3 x

i.e. 0. 55962 = 3 x

i.e. x=0.1865,

correct to 4 significant figures.

Problem 16. Solveex−^1 = 2 e^3 x−^4 correct to 4

significant figures

Taking natural logarithms of both sides gives

ln

(

ex−^1

)

=ln

(

2 e^3 x−^4

)

and by the first law of logarithms,

ln

(

ex−^1

)

=ln2+ln

(

e^3 x−^4

)

i.e. x− 1 =ln2+ 3 x− 4

Rearranging gives

4 − 1 −ln2= 3 x−x

i.e. 3 −ln2= 2 x

from which, x=

3 −ln2

2

=1.153

Problem 17. Solve, correct to 4 significant

figures, ln(x− 2 )^2 =ln(x− 2 )−ln(x+ 3 )+ 1. 6

Rearranging gives

ln(x− 2 )^2 −ln(x− 2 )+ln(x+ 3 )= 1. 6

and by the laws of logarithms,

ln

{

(x− 2 )^2 (x+ 3 )

(x− 2 )

}

= 1. 6

Cancelling gives

ln{(x− 2 )(x+ 3 )}= 1. 6

and (x− 2 )(x+ 3 )=e^1.^6

i.e. x^2 +x− 6 =e^1.^6

or x^2 +x− 6 −e^1.^6 = 0

i.e. x^2 +x− 10. 953 = 0

Using the quadratic formula,

x=

− 1 ±

√

12 − 4 ( 1 )(− 10. 953 )

2

=

− 1 ±

√

44. 812

2

=

− 1 ± 6. 6942

2

i.e. x= 2 .847 or− 3. 8471

x=− 3 .8471 is not valid since the logarithm of a

negative number has no real root.

Hence,the solution of the equation isx=2.847

Now try the following Practice Exercise

PracticeExercise 65 Evaluating Napierian

logarithms (answers on page 347)

In problems 1 and 2, evaluate correct to 5 signifi-

cant figures.

1. (a)

1

3

ln5. 2932 (b)

ln82. 473

4. 829

(c)

5 .62ln321. 62

e^1.^2942

2. (a)
   1 .786lne^1.^76
   lg10^1.^41

(b)

5 e−^0.^1629

2ln0. 00165

(c)

ln4. 8629 −ln2. 4711

5. 173

In problems 3 to 16, solve the given equations, each

correct to 4 significant figures.

3. 
   
   
   
   1. 5 = 4 e^2 t
   
   
   
   


4. 
   
   
   
   7. 83 = 2. 91 e−^1.^7 x
   
   
   
   


5. 16= 24

(

1 −e−

t

2

)

6. 
   
   
   
   5. 17 =ln
   
   
   
   

(

x

4. 64

)

7. 3.72ln

(

1. 59

x

)

= 2. 43

8. lnx= 2. 40


9. 24+e^2 x= 45


10. 5=ex+^1 − 7