Higher Engineering Mathematics, Sixth Edition

Logarithms 25

Rearranging gives

x=

log 1027

log 103

=

1. 43136 ...

0. 4771 ...

= 3

which may be readily checked
(

Note,

(

log8

log2

)

isnotequal to lg

(

8

2

))

Problem 22. 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 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 whichx=

2. 6865

0. 6532

=4.11,correct to

2 decimal places

Problem 24. 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 .2log 10 x=log 1041. 15

Hence log 10 x=

log 1041. 15

3. 2

= 0. 50449

Thusx=antilog0.50449= 100.^50449 =3.195correct to

4 significant figures.

Now try the following exercise

Exercise 13 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=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

when

n=25 decibels.

[316.2]

3.4 Graphs of logarithmic functions

A graph ofy=log 10 xis shown in Fig. 3.1 and a graph

ofy=logexis shown in Fig. 3.2. Both are 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 basea, it is noted

that:

(i) loga 1 = 0

Let loga=x,thenax=1 from the definition of

the logarithm.

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

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

that log 101 =0andloge 1 = 0