Leadership

Mathematics

Eleven Standard >> Logarithm | Introduction

Click the green "Start" button for MCQ.
Leadership

 

 Introduction of Logarithm

 

Let 'a' be a positive real number where a≠1 and M>0 is also a real number. Then there exists a real number 'x' which is called Logarithm of M to the base 'a' and denoted by \(\log_{a}{M}=x\) if and only if \(a^{x}=M\)

\(a^{x}=M\)  \(\Rightarrow \log_{a}{M}=x\)
Ex:
\(2^{5}=32\)  \(\Rightarrow \log_{2}{32}=5\)
\(10^{2}=100\)  \(\Rightarrow \log_{10}{100}=2\)

Some properties:

1) If a<0, a≠1 but M>0 then \(\log_{-2}{16}\) is not defined.

2) \(2^{4}=16\)
     \(\Rightarrow \log_{2}{16}=4\)
      \(\Rightarrow\)  \(4^{2}=16\)
      \(\Rightarrow \) \(\log_{4}{16}=2\)

Logarithm of different base may be different. For this reason, in order to find the logarithm of any positive real number, the base of the logarithm must be well defined.
  Any positive, fixed and finite real number a≠1 may be used as the base of the logarithm of any positive real number M


3) Logarithm of zero or a negative number is undefined.

\(2^{4}=16\)
\(\Rightarrow\) \(\log_{2}{16}=4\) ...(1)
\((-2)^{4}=16\)
\(\Rightarrow\) \(\log_{-2}{16}=4\) ....(2)
For this reason negative base is not allowed.

 

4) \((-1)^{1}=-1 \)
  \((-1)^{-1}=-1\)
     For this reason we consider only the positive real numbers neq ≠1 as the base of the logarithm of a real number (M>0)

Leadership
Hand drawn

Hide

Forgot your password?

Close

Error message here!

Hide

Lost your password? Please enter your email address. You will receive a link to create a new password.

Back to log-in

Close