Answer
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Hint: Logarithm of the form \[\log \,a\] has a base of the logarithm as 10. In logarithm we have several properties like \[\log \,\dfrac{m}{n}\,\,=\,\,\log m\,-\,\log n\] (where \[m\],\[n\] are positive numbers)
\[{{\log }_{a}}a\,\,=\,\,1\], \[\log 1\,=\,0\,\] etc.
Complete step by step solution:
Definition of logarithm:
Every positive real number \[N\] can be expressed in exponential form as \[{{a}^{x}}\,\,=\,\,N\]where ' \[a\]' is also a positive real number different than unity and is called the base and ' \[x\]' is called an exponent. We can write the relation \[{{a}^{x}}\,\,=\,\,N\]in logarithmic form as \[{{\log }_{a}}N\,=\,x\]. Hence \[{{a}^{x}}\,\,=\,\,N\,\,\Leftrightarrow \,\,{{\log }_{a}}N\,=\,x\].
Hence, the logarithm of a number to some base is the exponent by which the base must be raised in order to get that number.
Limitations of logarithm: \[{{\log }_{a}}N\,=\,x\] is defined only when (i)\[N\,>\,0\], (ii) \[a\,>\,0\] (iii) \[a\,\ne \,1\]
We can evaluate \[\log \,0.01\] step by step by transforming it in such a form so that we can apply the known formulae or properties of logarithm.
\[\log \,0.01\] can be written as \[\log \left( \dfrac{1}{100} \right)\].
As \[\log \,\dfrac{m}{n}\,\,=\,\,\log m\,-\,\log n\].
\[\,\Rightarrow \,\,\log \left( \dfrac{1}{100} \right)\,\,=\,\,\log 1\,-\,\log 100\,=\,0\,-\,\log {{10}^{2}}\] ……………………………………………………… (i)
also as \[\log 1\,=\,0\] and \[\log {{a}^{n}}\,=\,n\times \log a\](power rule of logarithm) equation (i) reduces to
\[\,\Rightarrow \,\,\log \left( \dfrac{1}{100} \right)\,\,=\,\,0\,\,-\,\,2\times \log (10)\,\,=\,\,0\,\,-\,\,2\times 1\]
\[\therefore \,\,\,\,\log 0.01\,\,=\,\,-2\].
Note:
> \[\log a\] has the base of the logarithm as 10 whereas \[\log a\] has the base of the logarithm as \[e\], where \[e\] is Napier’s constant. Napier’s constant is an irrational number. The approximate value of Napier’s constant is \[e\,\,=\,\,2.718\].
> For a given value of \[N\], \[{{\log }_{a}}N\] will give us a unique value.
> Logarithm of zero does not exist.
\[{{\log }_{N}}N\,=\,\,1\]
> Logarithms of negative real numbers are not defined in the system of real numbers.
\[{{\log }_{a}}a\,\,=\,\,1\], \[\log 1\,=\,0\,\] etc.
Complete step by step solution:
Definition of logarithm:
Every positive real number \[N\] can be expressed in exponential form as \[{{a}^{x}}\,\,=\,\,N\]where ' \[a\]' is also a positive real number different than unity and is called the base and ' \[x\]' is called an exponent. We can write the relation \[{{a}^{x}}\,\,=\,\,N\]in logarithmic form as \[{{\log }_{a}}N\,=\,x\]. Hence \[{{a}^{x}}\,\,=\,\,N\,\,\Leftrightarrow \,\,{{\log }_{a}}N\,=\,x\].
Hence, the logarithm of a number to some base is the exponent by which the base must be raised in order to get that number.
Limitations of logarithm: \[{{\log }_{a}}N\,=\,x\] is defined only when (i)\[N\,>\,0\], (ii) \[a\,>\,0\] (iii) \[a\,\ne \,1\]
We can evaluate \[\log \,0.01\] step by step by transforming it in such a form so that we can apply the known formulae or properties of logarithm.
\[\log \,0.01\] can be written as \[\log \left( \dfrac{1}{100} \right)\].
As \[\log \,\dfrac{m}{n}\,\,=\,\,\log m\,-\,\log n\].
\[\,\Rightarrow \,\,\log \left( \dfrac{1}{100} \right)\,\,=\,\,\log 1\,-\,\log 100\,=\,0\,-\,\log {{10}^{2}}\] ……………………………………………………… (i)
also as \[\log 1\,=\,0\] and \[\log {{a}^{n}}\,=\,n\times \log a\](power rule of logarithm) equation (i) reduces to
\[\,\Rightarrow \,\,\log \left( \dfrac{1}{100} \right)\,\,=\,\,0\,\,-\,\,2\times \log (10)\,\,=\,\,0\,\,-\,\,2\times 1\]
\[\therefore \,\,\,\,\log 0.01\,\,=\,\,-2\].
Note:
> \[\log a\] has the base of the logarithm as 10 whereas \[\log a\] has the base of the logarithm as \[e\], where \[e\] is Napier’s constant. Napier’s constant is an irrational number. The approximate value of Napier’s constant is \[e\,\,=\,\,2.718\].
> For a given value of \[N\], \[{{\log }_{a}}N\] will give us a unique value.
> Logarithm of zero does not exist.
\[{{\log }_{N}}N\,=\,\,1\]
> Logarithms of negative real numbers are not defined in the system of real numbers.
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