Higher Engineering Mathematics, Sixth Edition

Exponential functions 33

Taking natural logs of both sides gives:

ln

7

4

=lne^3 x

ln

7

4

= 3 xlne

Since lne=1ln

7

4

= 3 x

i.e. 0. 55962 = 3 x

i.e. x= 0. 1865 , correct to 4
significant figures.

Problem 15. Solve: ex−^1 =2e^3 x−^4 correct to 4

significant figures.

Taking natural logarithms of both sides gives:

ln

(

ex−^1

)

=ln

(

2e^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 16. 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 is:x=2.847

Now try the following exercise

Exercise 17 Further problemson

evaluating Napierian logarithms

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

[(a) 0.55547 (b) 0.91374 (c) 8.8941]

2. (a)

1 .786lne^1.^76

lg10^1.^41

(b)

5e−^0.^1629

2ln0. 00165

(c)

ln4. 8629 −ln2. 4711

5. 173

[(a) 2.2293 (b)− 0 .33154 (c) 0.13087]

In Problems 3 to 7 solve the given equations, each

correct to 4 significant figures.

3. lnx= 2. 10 [8.166]


4. 24+e^2 x= 45 [1.522]


5. 5=ex+^1 −7[ 1 .485]


6. 
   
   
   
   1. 5 =4e^2 t [− 0 .4904]
   
   
   
   


7. 
   
   
   
   7. 83 = 2 .91e−^1.^7 x [− 0 .5822]
   
   
   
   


8. 16= 24

(

1 −e−

t

2

)

[2.197]

9. 
   
   
   
   5. 17 =ln
   
   
   
   

( x

4. 64

)

[816.2]

10. 3.72ln

(

1. 59

x

)

= 2. 43 [0.8274]