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:
- 3x= 6. 4 [1.690]
- 2x= 9 [3.170]
- 2x−^1 = 32 x−^1 [0.2696]
- x^1.^5 = 14. 91 [6.058]
- 28 = 4. 2 x [2.251]
- 4^2 x−^1 = 5 x+^2 [3.959]
- x−^0.^25 = 0. 792 [2.542]
- 027 x= 3. 26 [−0.3272]
- 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