## Logarithms
Logarithms were invented independently by John Napier(in 1614) and Joost Burgi(in 1620). Log is being useful in many fields from astronomy to common finance.
We know multiplication is helpful shortcut for addition, and the topic exponent is one of the short cut for multiplication, and log is the short cut for exponents.
Let b a positive number b≠ 1.For any real number x, b^{xx}, then the exponent x is called the log of a to the base b. We also call x, the value of log, log_{ b} a. Thus x = log_{b} a is an equivalent form of a= b^{x}. We say that x = log_{b} a is the logarithmic form of the exponential form a = b^{x}. In both forms the base is same. We observe that x = log_{b} a is an equivalent way of writing a = b^{x}.
**For example,**
- 3 = log
_{9} 729 is equivalent to 9^{3} = 729.
- 1/3 = log
_{8} 2 is equivalent to 8^{1/3} = 2.
- -3 = log
_{10} 0.001 is equivalent to 10^{-3} = 0.001.
The Base must be specified in the logrithmic notation. If we write y= log x, then it will become meaningless since its equivalent cannot be written unless the base is given. However, in such situations, we write logarithm, by omitting their base. In such case, it is understood that all log is having the same base.
**Change the following logrithmic form into exponential form:**
** (i) log**_{25} 5 = 1/2
**Solution:**
5=25^{1/2}
** (ii)log**_{216} 6 = 1/3
**Solution:**
6=216^{1/3}.
**(iii) log**_{1/4} =-2
**Solution:**
1/2=2^{-2}.
**Change the following exponential form into logrithmic form:**
** (i) 2=64**^{1/6}
**Solution:**
log_{64} 2= 1/6
**(ii) 9**^{-3 = 1/729}.
**Solution:**
-3 log_{9} [1/729]
Evaluate:
(i) log_{9} 729
**solution:**
Let x = log_{9} 729
9^{x}=729
We know that 729 = 9x9x9=9^{3}
So 9^{x}= 9^{3}
x=3
We can evaluate this using laws of logrithm,which we are going to see in the upcoming pages.
**(ii) log**_{4} 8
**Solution:**
Let x=log_{4}8
4^{x}=8
As we know that 4=2^{2} and 8=2^{3}
2^{2x}=2^{3}
2x=3
x=3/2
**(iii) log**_{9} 1/27
**Solution:**
Let x= log_{9} 1/27
9^{x}=1/27
As we know that,
9=3^{2}and 27=3^{3}
3^{2x}=1/(3^{3})
3^{2x}=3^{-3}
which implies that 2x=-3
**x=-3/2**
Solve the equations:
(i) log_{3}x=-2.
**Solution:**
log_{3}x=-2
x=3^{-2}
x=1/9
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Rules of Logarithms
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