How do you evaluate ${{\log }_{5}}\left( \dfrac{1}{25} \right)$.

Hint: In the question we have logarithmic function with base $5$. So, it can be easier when we convert the numerical part into the exponential of $5$. Because we have a logarithmic law ${{\log }_{a}}a=1$. So first we will convert the numerical part which is $\dfrac{1}{25}$ into the power of $5$. For this we will use the well known relation $25={{5}^{2}}$. We will substitute this value in the given equation and now we will apply the exponential rules which is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. After that we will use the logarithmic formula $\log {{a}^{m}}=m\log a$. Finally, we can apply the formula ${{\log }_{a}}a=1$ to get the result.

Complete step by step answer:
Given that, ${{\log }_{5}}\left( \dfrac{1}{25} \right)$.
Considering the numerical part separately, then the numerical part is $\dfrac{1}{25}$.
In denominator we have the value $25$. We can write $25$ as ${{5}^{2}}$. So, substituting the value $25={{5}^{2}}$ in the denominator of the numerical part, then we will have
$\dfrac{1}{25}=\dfrac{1}{{{5}^{2}}}$
We have the exponential rules which is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. Applying this formula in the above equation, then we will have
$\dfrac{1}{25}={{5}^{-2}}$.
Applying the above value in the given logarithmic function, then we will gat
${{\log }_{5}}\left( \dfrac{1}{25} \right)={{\log }_{5}}\left( {{5}^{-2}} \right)$
We have the logarithmic formula $\log {{a}^{m}}=m\log a$. Applying this formula in the above equation, then we will have
$\Rightarrow {{\log }_{5}}\left( \dfrac{1}{25} \right)=-2{{\log }_{5}}5$
Applying the formula ${{\log }_{a}}a=1$ in the above equation, then we will have
$\begin{align}
  & \Rightarrow {{\log }_{5}}\left( \dfrac{1}{25} \right)=-2\left( 1 \right) \\
 & \therefore {{\log }_{5}}\left( \dfrac{1}{25} \right)=-2 \\
\end{align}$

Note:
In logarithmic we have different formulas that will help in solving this kind of problems. Some of the formulas which we will regularly use are listed below.
$\log \left( ab \right)=\log a+\log b$
$\log \left( \dfrac{a}{b} \right)=\log a-\log b$
The above formulas are used very regularly according to the problem.