Basic Engineering Mathematics, Fifth Edition

116 Basic Engineering Mathematics

Problem 22. Solve the equation 2x=5, correct

to 4 significant figures

Taking logarithms to base 10 of both sides of 2x= 5

gives

log 102 x=log 105

i.e. xlog 102 =log 105

by the third law of logarithms

Rearranging givesx=

log 105

log 102

=

0. 6989700 ...

0. 3010299 ...

=2.322,correct to 4

significant figures.

Problem 23. 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 =( 2 x− 5 )log 103

xlog 102 +log 102 = 2 xlog 103 −5log 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 which x=^2.^6865

0. 6532

=4.11,

correct to 2 decimal places.

Problem 24. Solve the equationx^2.^7 = 34 .68,

correct to 4 significant figures

Taking logarithms to base 10 of both sides gives

log 10 x^2.^7 =log 1034. 68

2 .7log 10 x=log 1034. 68

Hence, log

10 x=

log 1034. 68

2. 7

= 0. 57040

Thus, x=antilog 0. 57040 = 100.^57040

=3.719,

correct to 4 significant figures.

Now try the following Practice Exercise

PracticeExercise 61 Indicial equations

(answers on page 346)

In problems 1 to 8, solve the indicial equations for

x, each correct to 4 significant figures.

1. 3x= 6 .42.2x= 9


2. 2x−^1 = 32 x−^1 4. x^1.^5 = 14. 91


3. 
   
   
   
   25. 28 = 4. 2 x 6. 4^2 x−^1 = 5 x+^2
   
   
   
   


4. x−^0.^25 = 0 .792 8. 0. 027 x= 3. 26
   9. The decibel gainnof an amplifier is given
   byn=10log 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

whenn=25 decibels.

15.4 Graphs of logarithmic functions

A graph ofy=log 10 xis shown in Figure 15.1 and a

graph ofy=logexis shown in Figure 15.2. Both can

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

shape occurs for a logarithm to any base.

In general, with a logarithm to any base,a, it is noted

that

(a) loga 1 = 0

Let loga=xthenax=1 from thedefinition of the

logarithm.

Ifax=1thenx=0 from the laws of logarithms.

Hence,loga 1 = 0. In the above graphs it is seen

that log 10 1=0andloge 1 =0.

(b) logaa= 1

Let logaa=xthenax=afrom the definition of

a logarithm.

Ifax=athenx=1.