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We know, 23 8 ; this mathematical statement is written in terms of log as

log 28 3. Again, conversely, if log 28 3 , it can be written in terms of exponents

as 23 8. That is, if 23 8 , then log 28 3 and conversely, if log 28 3 , then

23 8. Similarly,

8

1

2

1

2 ^3 3 can be written in terms of log as 3

8

1

log 2 .

If ax N, (a! 0 ,az 1 ),x logaN is defined as a based logN.
To be noticed : Whatever may be the values of x, positive or negative, ax is
always positive. So, only the log of positive numbers has values which are real ; log
of zero or negative numbers have no real value.

Activity-1 : Express in terms of

log :

Activity-2 : Fill in the blanks :

(i) 102 100 in terms of exponent in terms of log

(ii)

9

1

3 ^2  101

(^0) log 101 0
(iii)
2
1
22
1
 e^0 ..... loge 1 ......
(iv) 24 1
a^0 ...... = ......
101 10 log 1010 1
e^1 ... ....... = .......
....... = ...... logaa 1
Formulae of Logarithms :
Let, a! 0 ,az 1 ;b! 0 ,bz 1 andM! 0 ,N! 0.
Formula 1. (a) loga 1 0 ,(a! 0 ,az 1 )
(b) logaa 1 ,(a! 0 ,az 1 )
Proof : (a) We know from the formula of exponents, a^0 1
? from the definition of log, we get, loga 1 0 (proved )
(b) We know, from the formula of exponents, a^1 a
? from the definition of log, we get, logaa 1 (proved).
Formula 2. loga(MN) logaMlogaN
Proof : Let, logaM x,logaN y;
? M ax,N ay