2.3. Logarithms:
In 2.2 we learnt exponents and several laws.
To recollect, some numbers can be represented as:
Base ^{Exponent }= Number
This method of representation is called ‘exponential form’.
Ex: 2^{3 }= 8
This relationship can be represented in a different
format:
log _{base
}Number = Exponent
EX : log_{2 }8 = 3 ( Equivalent to
saying 8 = 2^{3})
The ‘logarithm’ (log)
of any number to a given base is the value of
the index to which the base must be raised
to get the given number
The above method of representation is called ‘logarithmic form’.
We have seen that
1) x^{0 }=1
Hence log_{x}1 = 0
2) x^{1 }=x
Hence log_{x}x = 1
From laws of exponents we can derive the following
laws of logarithms:
1. Product law:
log_{a}(mn) = log_{a }m+ log_{a }n (By x^{m}.x^{n}= x^{m+n}
property)
2. Quotient law: log_{a}(m/n) = log_{a }m
 log_{a }n(By x^{m}/x^{n}=
x^{mn} property)
3. Power law:
log_{a}m^{n} = nlog_{a }m(By (x^{m})^{n}= x^{mn} property)
4. log_{a}m^{1/n } = (1/n) log_{a}m ( By Power law)
2.3 Problem 1: Find x if log_{9} 243=x
Solution:
This is equivalent to saying 9^{x}= 243 = 3*3*3*3*3 = 3^{5}
But 9^{x}=3^{2x}
_{}3^{2x}= 3^{5}
_{} 2x =5
_{} x =5/2^{}
_{ }
Logarithms to the base 10
are called ‘common logarithms’
Convention: If base is not given, then the
base is assumed to be 10 (log20 implies log_{10}20).
We have seen that log1=0 and log10=1.
2.3 Problem 2: Given log x = m+n and log y = mn. Express log (10x/y^{2})
in terms of m and n.
Solution:
log (10x/y^{2}) = log 10x – log y^{2 }(quotient
law)
= log 10 + log x – 2 log y( product law, power law)
= 1 + (m+n) – 2(mn) (substitution, log 10 = 1)
= 1m+3n
2.3 Problem 3: Given 2log x+1 = log 250 find x, log 2x
Solution:
Since base is not
given, it is assumed to be 10.
log 250 = log (10*25) = log 10 + log 25 = 1 + log 25
(log 10 = 1)
_{} 2log x + 1 = 1 + log
25
i.e. 2log x = log 25
But 2log x = log x^{2}(Power law)
_{} log x^{2 }= log 25
_{} x^{2}=25^{}
i.e. x=5^{}
_{} log 2x = log 10 = 1
2.3 Problem 4: If 2log y – log x – 3 = 0 express x in terms of y
Solution:
Since 2log y – log x –
3 = 0
log x = 2log
y  3 = 2log y – 3log 10
= log y^{2} log
10^{3} = log (y^{2}/10^{3})
Thus
log x = log (y^{2}/10^{3})
_{} x = y^{2}/10^{3} = y^{2}/1000
2.3 Problem 5: If log 2 = 0.3010 and log 3 = 0.4771 find value of log _{}
Solution:
_{} = _{}= (30/4)^{1/2}
_{} log _{} = (1/2)log(30/4) =
1/2(log30log4)
= ˝(log 10 + log 3 
2log 2)
= ˝(1+0.47710.6020)
=˝(0.8751)
= 0.43755^{}
We have seen that log 1 = 0, log 10 = 1(_{}log_{x }x = 1 and log_{x }1 = 0)
Thus log x for any value of x (1_{} x_{} 10) has to be in between 0 and 1.
In fact log 2 = 0.3010
and log 3 = 0.4771 (We get these values from log tables discussed later)
Also note
log 100 = log(10*10)=log10+log10=2log10=2
log 1000 = log(10^{3}) = 3log10=3*1=3 and
so on….
log value of a number
has two parts: an integer part and decimal part.
The integral part of the logarithm is called ‘characteristic’ and the decimal
part of the logarithm is called ‘mantissa’.
In case of log 2 = 0.3010, 0 is characteristic and .3010
is mantissa.
We use a table called ‘logarithmic table’ to find the values of numbers
to the base 10.
A section of the table
is given below:
As we observe from adjacent
figure, the table has three parts. 1. First part is the left most
column (Blue
box) which contains numbers
from 1.0 to 1.5(The adjacent figure only has a part
of the table from 1.0 to 1.5 but the full table has values from 1.0 to 9.9). 2. Second part contains 10
columns (red
box) 3. Third part contains 9 columns (pink box)
(with Mean differences as the heading) Against each number in the first
column there are 10 columns in 2nd part,
having decimal values corresponding to digits from 0 to 9. Against each number in the first
column there are 9 columns in 3rd part, having decimal values corresponding
to digits from 1 to 9. 

If we look at the full
logarithmic table we notice the values
are from 0.0000 to 1.0 (0.9996+.0004) The table represents values from
0 to 1 correspondingly from log1 to log10 (log 1 = 0, log 10 = 1) 
Observe the following:
Number 
=(Equivalent) 
log of number 
= 
=(Expansion) 
=(Simplification) 
3257 = ^{} 
3.257*10^{3} 
log(3257) = ^{} 
Log(3.257*10^{3}) 
= 3log10 +log(3.257) 
= 3+ log(3.257) 
325.7 = 
3.257*10^{2} 
log(325.7) = 
Log(3.257*10^{2}) 
= 2log10 +log(3.257) 
= 2+ log(3.257) 
32.57 = ^{} 
3.257*10^{1} 
log(32.57) = ^{} 
Log(3.257*10^{1}) 
= 1log10 +log(3.257) 
= 1+ log(3.257) 
3.257= 
3.257^{} 
log(3.257)= 
Log(3.257) 
= 0log10 + log(3.257) 
= log(3.257) 
0.3257 = ^{} 
3.257*10^{1} 
log(0.3257) = ^{} 
Log(3.257*10^{1}) 
= 1log10 +log(3.257) 
= _{}+log(3.257) 
0.03257 = 
3.257*10^{2} 
log(0.03257) = 
Log(3.257*10^{2}) 
= 2log10+log(3.257) 
= _{}+log(3.257) 
0.003257 = 
3.257*10^{3} 
log(0.003257) = 
Log(3.257*10^{3}) 
= 3log10+log(3.257) 
= _{}+log(3.257) 
0.0003257 = 
3.257*10^{4} 
log(0.0003257) = 
Log(3.257*10^{4}) 
= 4log10+log(3.257) 
= _{}+log(3.257) 
Note: –log10 is not written
as –1 but as _{}
Thus if we know the logarithmic
value of a number having one integer digit, then by using product and power
laws of logarithm we can find the values of any given number which is _{} multiple of the given
number.
Note that the
logarithmic table given in the adjacent figure
is for values from log
1.0 to log(3.5)
(Note: log 1 = 0)
How to use the log table?
Let us find the value of log
32.57 1)First convert the given number into a decimal such that there
is only one digit in the integer part and there are not more than three
digits in the decimal part. Thus 32.57 = 3.257*10^{1}_{}log(32.57) = 1+ log(3.257) ^{ } 2)Split the given four digit number into 3 parts (first
part with two digits, then two parts of one digit each).In this example 32.57
is split into three parts (3.2, 5 and 7). 3)Identify the row corresponding to first part of 2
numbers (In this example 3.2). 4)In this row find the value under the number which corresponds
to the second part(in this example under 5).
The value is 0.5119 5)To the above value, add the value under the
number which corresponds to the third part (in this example under 7). The value is 0.0009 6)Sum of values of steps 4 and 5 is the log value of 1/10^{th}
of the given number(32.57)(0.5128 = 0.5119 + 0.0009) is the log value of 3.257 _{}log(32.57) = 1+ log(3.257) = 1+0.5128 Similarly log (2.223) = 0.3464+0.006
= 0.3470 and hence log(22.23) = 1.3470, log(222.3) = 2.3470,…… 

2.3 Problem 6: Find a, b and c If log(a) =
1.7782, log(b)=
_{}.7782 and log(c)=1.5564 and log6=0.7782
Solution:
Log(a)=1.7782 = 1+0.7782 = log10+log6 = log60
_{}a=60
log(b)= _{}.7782 = log(10^{2})+0.7782 = log10^{2}+log6
= log(6/100)
_{}b=0.06
log(c)=1.5564 = 2 (0.7782) {By intuition} = 2 log6 = log 6^{2}
_{}c=6^{2}=36
If log3257 = 3.5128,
then 3257 is called ‘antilogarithm’ of
3.5128 and we write antilog3.5128 = 3257 or log^{1 }(3.5128) = 3257.
Like logarithmic tables
we also have antilog tables, a section of which is given below. The usage of
antilog table is similar to log table.
If log table has values
corresponding to 1.00 – 9.9, antilog table has values corresponding to 0.00 to
.99
Change of
base of logarithm:
log_{x }y= log_{x }a/ log_{y }a
Proof:
Let log_{x }a = p,
_{}x^{p}=a ŕ(1)
Let log_{y
}a=q
_{} y^{q}=a ŕ(2)
log_{x }y = r
And Let log_{x }y =r
_{} x^{r}=y ŕ(3)
From (2) a = y^{q }( by substituting from (3) for y) we get^{}
a=( x^{r})^{q}= x^{rq} (By
law of exponents)
But from (1) a = x^{p}
Hence x^{p}= x^{rq}
_{} p = qr
By substituting values for p, q and r.
log_{x }a = log_{y }a * log_{x }y
i.e. log_{x }y = log_{x }a/ log_{y
}a
Similarly we can prove that
log_{x }y = log_{a }y/ log_{a }x
2.3 Problem 7: If x = 1+log_{a}bc, y= 1+log_{b}ca , z =
1+log_{c}ab, show that xy+yz+zx= xyz
Solution:
Since x = 1+log_{a}bc,
log_{a}bc =x1
_{}bc = a^{x1}
= a^{x}/a
_{}abc = a^{x}
_{} (abc)^{1/x} =
a
Similarly
(abc)^{1/y} =b
and (abc)^{1/z }=c
_{}(abc)^{1/x}(abc)^{1/y} (abc)^{1/z}= abc
(By multiplying individual terms)
But
(abc)^{1/x}(abc)^{1/y} (abc)^{1/z} = (abc)^{(1/x+1/y+1/z)}
_{}abc = (abc)^{(1/x+1/y+1/z)}
_{} 1 =
[(1/x)+(1/y)+(1/z)] ( By equating exponents)
=
(yz/xyz)+(xz/xyz)+(xy/xyz)
= (xy+yz+zx)/xyz
_{} xyz = xy+yz+zx
Why use Logarithmic and antilogarithmic tables?
2.3 Summary of
learning
No 
Points to remember 
1 
log_{a}(mn) = log_{a }m+ log_{a }n 
2 
log_{a}(m/n) = log_{a }m – log_{a }n 
3 
log_{a}m^{n}
= nlog_{a }m 
4 
log_{a}m^{1/n
} = (1/n) log_{a}m 
5 
log_{x
}y= log_{x }a/ log_{y }a 
6 
log_{x
}y = log_{a }y/ log_{a }x 