How do you evaluate \[{{\log }_{4}}\left( \dfrac{1}{2} \right)\]?

Hint: First of all convert the argument of the given logarithmic expression into exponential form by using the formula: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\]. Now, use the identity: - \[\log {{a}^{m}}=m\log a\] for the simplification of the expression. Now, convert the base of the logarithmic into exponential form by writing \[4={{2}^{2}}\] and use the identity: - \[{{\log }_{{{a}^{b}}}}m=\dfrac{1}{b}{{\log }_{a}}m\] for further simplification. Finally, use the formula: - \[{{\log }_{n}}n=1\] to get the answer. Here, n > 0 and \[n\ne 1\].

Complete answer:
Here, we have been provided with the logarithmic expression \[{{\log }_{4}}\left( \dfrac{1}{2} \right)\] and we have been asked to evaluate it. That means we have to find its value.
Now, let us assume the value of the given expression is ‘E’. So, we have,
\[\Rightarrow E={{\log }_{4}}\left( \dfrac{1}{2} \right)\]
Here, we have the argument of log equal to \[\dfrac{1}{2}\], so writing this fraction in exponential form using the formula: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\], we have,
\[\Rightarrow E={{\log }_{4}}\left( {{2}^{-1}} \right)\]
Using the formula: - \[\log {{a}^{m}}=m\log a\], we get,
\[\begin{align}
  & \Rightarrow E=\left( -1 \right)\times {{\log }_{4}}2 \\
 & \Rightarrow E=-{{\log }_{4}}2 \\
\end{align}\]
Now, as we can see that the base of this logarithmic expression is which can be converted into exponential form by using the prime factorization. So, we can write,
\[\Rightarrow 4=2\times 2\]
This can be written as: -
\[\Rightarrow 4={{2}^{2}}\]
Therefore, the expression becomes,
\[\Rightarrow E=-{{\log }_{{{2}^{2}}}}2\]
Using the formula: - \[{{\log }_{{{a}^{b}}}}m=\dfrac{1}{b}{{\log }_{a}}m\], we get,
\[\Rightarrow E=\dfrac{-1}{2}{{\log }_{2}}2\]
As we can see that the argument and base of the logarithm has become equal so we can use the formula: - \[{{\log }_{n}}n=1\] for n > 0 and \[n\ne 1\]. Therefore, the expression ‘E’ can be simplified as: -
\[\begin{align}
  & \Rightarrow E=\left( \dfrac{-1}{2} \right)\times 1 \\
 & \Rightarrow E=\left( \dfrac{-1}{2} \right) \\
\end{align}\]
Hence, the value of the given logarithmic expression is \[\left( \dfrac{-1}{2} \right)\].

Note:
You must remember the basic formulas of logarithm like: - \[\log m+\log n=\log \left( mn \right)\], \[\log m-\log n=\log \left( \dfrac{m}{n} \right)\], \[\log {{m}^{n}}=n\log m\], \[{{\log }_{{{a}^{b}}}}m=\dfrac{1}{b}{{\log }_{a}}m\] etc. You may note the exponentiation is the reverse process of logarithm so we can also solve this question by converting the logarithmic form of equation into exponential form by using the formula: - if \[x={{\log }_{a}}m\] then \[m={{a}^{x}}\]. In the next step we will write: - \[{{4}^{x}}=\dfrac{1}{2}\], which can be further simplified as \[{{4}^{x}}={{2}^{-1}}\]. Now, we will compare \[{{2}^{2x}}={{2}^{-1}}\] and then equate the exponents (-1) and (2x) to solve for the value of x.