How do you evaluate \[\log _{2}^{{}}\left( \dfrac{1}{64} \right)\]?

Hint: These types of problems can be solved by considering the solution as any variable. Then consider the equation as equation (1). Now we have to apply the basic logarithm formula and then apply this formula to equation (1). Now, by applying some exponential formulas we will get the solution of the problem.

Complete step by step answer:
From the given question, we are given to solve\[\log _{2}^{{}}\left( \dfrac{1}{64} \right)\]
Now let us assume the given equation as ‘a’.
\[a=\log _{2}^{{}}\left( \dfrac{1}{64} \right)\]
For solving the equation \[a=\log _{2}^{{}}\left( \dfrac{1}{64} \right)\] let us consider it as equation (1).
\[a=\log _{2}^{{}}\left( \dfrac{1}{64} \right)..................(1)\]
As we know that
\[\begin{align}
  & if\text{ }{{\log }_{a}}N=x; \\
 & then\text{ }{{a}^{x}}=N. \\
\end{align}\]
Now we have to apply the above concept to the equation (1).
By applying the above concept, we get
\[{{2}^{a}}=\dfrac{1}{64}\]
Let us consider the above equation as equation (2).
\[{{2}^{a}}=\dfrac{1}{64}..........\left( 2 \right)\]
Now we have to evaluate the above equation and we have to find the value of a.
We can write \[64\] as \[{{2}^{6}}\]. So, now substitute this in equation (2).
\[{{2}^{a}}=\dfrac{1}{{{2}^{6}}}\]
Let us consider the above equation as equation (3).
\[{{2}^{a}}=\dfrac{1}{{{2}^{6}}}........\left( 3 \right)\]
By the formula
\[\dfrac{1}{a}={{a}^{-1}}\].
By applying the above concept to equation (3), we get
\[{{2}^{a}}={{2}^{-6}}\]
Let us consider the above equation as equation (4)
\[{{2}^{a}}={{2}^{-6}}...........\left( 4 \right)\]
By the formula
\[\begin{align}
  & if\text{ }{{a}^{m}}={{a}^{n}} \\
 & then\text{ }m=n \\
\end{align}\]
By applying the above concept to the equation (4), we get
\[a=-6\]
Let us consider the above equation as equation (5).
\[a=-6.......(5)\]

So, therefore by solving \[a=\log _{2}^{{}}\left( \dfrac{1}{64} \right)\] we will get \[a=-6\].

Note: Some students may have a misconception that
\[\begin{align}
  & if\text{ }{{\log }_{a}}N=x; \\
 & then\text{ }{{a}^{x}}=N \\
\end{align}\].
But we know that
\[\begin{align}
  & if\text{ }{{\log }_{a}}N=x; \\
 & then\text{ }{{a}^{x}}=N. \\
\end{align}\] .
 If this misconception is followed, then the final answer may get interrupted. So, these misconceptions should be avoided. Also students should avoid calculation mistakes while solving the problem.