Question

How do you calculate ${{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)$ ?

Answer 1

Hint: We have been given a logarithmic function whose base is equal to 11 and whose argument is in a fractional format with numerator equal to 1 and denominator equal to square root of 11. In order to simplify and solve this complex expression, we shall use various properties of the logarithmic functions out of which the most essential one would be that the logarithm function would be equal to 1 when it has the same base and argument.

Complete step by step solution:
Given that ${{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)$.
When we have to write the term in the denominator of a fraction in its numerator, then we assign a power of -1 to that term in the denominator. We shall make this transition to the term in the denominator which is square root of 11 in order to write it in the numerator.
$\Rightarrow {{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)={{\log }_{11}}{{\left( \sqrt{11} \right)}^{-1}}$
Also, we will remove the square root by assigning a power of $\dfrac{1}{2}$ to 11.
$\Rightarrow {{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)={{\log }_{11}}{{\left( 11 \right)}^{-1\times \dfrac{1}{2}}}$
$\Rightarrow {{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)={{\log }_{11}}{{\left( 11 \right)}^{-\dfrac{1}{2}}}$
Another property of logarithmic functions says that when an exponential term is given in the logarithmic function, then the power can be taken outside the function and entire function is modified as the product of the power of the argument and the logarithm function of base of exponent, that is, $\log {{a}^{b}}=b\log a$.
Applying this property, we get
$\Rightarrow {{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)=-\dfrac{1}{2}.{{\log }_{11}}\left( 11 \right)$
Since the base and argument of the logarithmic function is the same and is equal to 11 now, thus we shall write it equal to 1.
$\Rightarrow {{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)=-\dfrac{1}{2}.\left( 1 \right)$
$\Rightarrow {{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)=-\dfrac{1}{2}$
Therefore, ${{\log }_{11}}\left( \dfrac{1}{\sqrt{11}} \right)$ is equal to $-\dfrac{1}{2}$.

Note: Another important property of the logarithmic functions is that when a logarithm function is expressed in terms of the product of two terms, then that logarithm function can be expressed as the sum of two individual logarithmic functions of those two terms, that is, $\log ab=\log a+\log b$.