# LOGARITHM

Logarithm of a number to a given base is the index or the power to which the base must be raised to produce the number. That is, to make it equal to the given number.

Let us consider the three quantities, sat a, x and n and they can be related as follows.

If ax  =  n, where n > 0, a > 0 and a ≠ 1, then x is said to be the logarithm of the number n to the base 'a',  symbolically it can be expressed as follows.

logan  =  x

Here, 'n' is called as argument.

## Fundamental Laws of Logarithm

Law 1 :

Logarithm of product of two numbers is equal to the sum of the logarithms of the numbers to the same base.

That is,

logamn  =  logam + logan

Law 2 :

Logarithm of the quotient of two numbers is equal to the difference of their logarithms to the same base.

That is,

loga(m/n)  =  logam - logan

Law 3 :

Logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number to the same base.

That is,

logamn  =  nlogam

## Relationship between Indices and Logarithms

Example 1 :

If logam  =  x, then

m  =  ax

Example 2 :

If k  =  xy, then

logxk  =  y

## Multiplication of Logarithms

Multiplication of two or more logarithms can be simplified, if one of the following two conditions is met.

Condition (i) :

The argument of the first logarithm and the base of the second logarithm must be same.

Condition (ii) :

The base of the first logarithm and the argument of the second logarithm must be same.

Let us see how the multiplication of two logarithms can be simplified in the following examples.

Example 1 :

Simplify : loga⋅ logbc

In the above two logarithms, the argument of the first logarithm and the base of the second logarithm are same.

So, we can simplify the multiplication of above two logarithms by removing the part circled in red color. Example 2 :

Simplify : logx⋅ logzx

In the above two logarithms, the base of the first logarithm and the argument of the second logarithm are same.

So, we can simplify the multiplication of above two logarithms by removing the part marked in red color. ## Change of Base

If the logarithm of a number to any base is given, then the logarithm of the same number to any other base can be determined from the following relation.

logba  =  logx⋅ logbx

logba  =  logxa / logxb

More clearly, In the above example, the base of the given logarithm 'b' is changed to 'x'.

In this way, we can change the base of the given logarithm to any other base.

## Inverse Properties of Logarithms ## Some other Properties of Logarithms

Property 1 :

If you switch a logarithm from numerator to denominator or denominator to numerator, we have to interchange the argument and base.

Examples :

logab  =  1 / logba

5 / (2logxy)  =  5logyx / 2

Property 2 :

Logarithm of any number to the same base is equal to 1.

Examples :

logxx  =  1

log33  =  1

The above property will not work, if both the base and argument are 1.

That is,

log11  ≠  1

The reason for the above result is explained in the next property.

Property 3 :

Logarithm of 1 to any base is equal to 0.

Examples :

logx1  =  0

log31  =  0

log11  =  0

## Notes

1.  If the base is not given in a logarithm, it can be taken as 10.

Thus,

log10  =  log1010  =  1

log1  =  log101  =  0

2.  Logarithm using base 10 is called Common logarithm and logarithm using base 'e' is called Natural logarithm.

Here, e    2.718 called exponential number.

## Practice Problems

Problem 1 :

Find the logarithm of 64 to the base 2√2.

Solution :

Write 64 as in terms of 2√2.

64  =  26

64  =  24+2

64  =  2 22

64  =  2⋅ [(√2)2]2

64  =  2⋅ (√2)4

64  =  (2√2)4

Then,

log2√264  =  log2√2(2√2)4

log2√264  =  4log2√2(2√2)

log2√264  =  4(1)

log2√264  =  4

Problem 2 :

If logabc  =  x, logbca  =  y and logcab  =  z, then find the value of

1/(x+1) + 1/(y+1) + 1/(z+1)

Solution :

x + 1  =  logabc + logaa  =  logaabc

y + 1  =  logbca + logbc  =  logbabc

z + 1  =  logcab + logcc  =  logcabc

1/(x+1)  =  1 / logaabc  =  logabca

1/(y+1)  =  1/logbabc  =  logabcb

1/(z+1)  =  1/logcabc  =  logabcc

1/(x+1) + 1/(y+1) + 1/(z+1)  =  logabca + logabcb + logabcc

1/(x+1) + 1/(y+1) + 1/(z+1)  =  logabcabc

1/(x+1) + 1/(y+1) + 1/(z+1)  =  1

Problem 3 :

If a = log2412, b = 2436 and c = log4836, then find the value of

1 + abc

Solution :

1 + abc  =  1 + log2412  log3624 ⋅ log4836

1 + abc  =  1 + log3612 ⋅ log4836

1 + abc  =  1 + log4812

1 + abc  =  log4848 + log4812

1 + abc  =  log48(48 ⋅ 12)

1 + abc  =  log48(2 ⋅ 12)2

1 + abc  =  2log4824

1 + abc  =  2log3624 ⋅ log4836

1 + abc  =  2bc Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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