Higher Engineering Mathematics

LOGARITHMS AND EXPONENTIAL FUNCTIONS 27

A

Problem 8. Solve the equation 2x=3, correct

to 4 significant figures.

Taking logarithms to base 10 of both sides of 2x= 3

gives:

log 102 x=log 103

i.e. xlog 102 =log 103

Rearranging gives:

x=

log 103

log 102

=

0. 47712125 ...

0. 30102999 ...

=1.585, correct to 4 significant figures

Problem 9. Solve the equation 2x+^1 = 32 x−^5

correct to 2 decimal places.

Taking logarithms to base 10 of both sides gives:

log 102 x+^1 =log 1032 x−^5

i.e. (x+1) log 102 =(2x−5) log 103

xlog 102 +log 102 = 2 xlog 103 −5 log 103

x(0.3010)+(0.3010)= 2 x(0.4771)−5(0.4771)

i.e. 0. 3010 x+ 0. 3010 = 0. 9542 x− 2. 3855

Hence

2. 3855 + 0. 3010 = 0. 9542 x− 0. 3010 x

2. 6865 = 0. 6532 x

from whichx=

2. 6865

0. 6532

=4.11, correct to

2 decimal places

Problem 10. Solve the equationx^3.^2 = 41 .15,

correct to 4 significant figures.

Taking logarithms to base 10 of both sides gives:

log 10 x^3.^2 =log 1041. 15

3 .2 log 10 x=log 1041. 15

Hence log 10 x=

log 1041. 15

3. 2

= 0. 50449

Thus x=antilog 0.50449= 100.^50449 =3.195 cor-

rect to 4 significant figures.

Now try the following exercise.

Exercise 17 Indicial equations

Solve the following indicial equations forx, each

correct to 4 significant figures:

1. 3x= 6. 4 [1.690]


2. 2x= 9 [3.170]


3. 2x−^1 = 32 x−^1 [0.2696]

4.x^1.^5 = 14. 91 [6.058]

5. 
   
   
   
   25. 28 = 4. 2 x [2.251]
   
   
   
   


6. 4^2 x−^1 = 5 x+^2 [3.959]

7.x−^0.^25 = 0. 792 [2.542]

8. 
   
   
   
   0. 027 x= 3. 26 [−0.3272]
   
   
   
   


9. The decibel gainnof an amplifier is given by:

n=10 log 10

(

P 2

P 1

)

whereP 1 is the power input andP 2 is the

power output. Find the power gain

P 2

P 1

when

n=25 decibels.

[316.2]

4.4 Graphs of logarithmic functions

A graph ofy=log 10 xis shown in Fig. 4.1 and a

graph ofy=logexis shown in Fig. 4.2. Both are

seen to be of similar shape; in fact, the same general

shape occurs for a logarithm to any base.

Figure 4.1