How to solve $x:{{\log }_{10}}\left( \dfrac{10x}{3} \right)-2={{\log }_{10}}\left( \dfrac{1}{10} \right)$ ?

Hint: Here in this question we have been asked to solve the given logarithmic expression ${{\log }_{10}}\left( \dfrac{10x}{3} \right)-2={{\log }_{10}}\left( \dfrac{1}{10} \right)$ for $x$ . For that sake we will use the following logarithmic identities.
${{\log }_{a}}{{a}^{p}}=p$
Here $a$ is assumed to be the base.

Complete step by step solution:
Now considering from the question we have been asked to solve the given logarithmic expression ${{\log }_{10}}\left( \dfrac{10x}{3} \right)-2={{\log }_{10}}\left( \dfrac{1}{10} \right)$ for $x$ .
From the basic concepts we know that the definition of logarithm is given as ${{\log }_{a}}x=y\Rightarrow {{a}^{y}}=x$ .
For that sake we will use the following logarithmic identities.
${{\log }_{a}}{{a}^{p}}=p$ here $a$ is assumed to be the base.
By using the logarithmic definition we can prove this formula.
By using ${{\log }_{a}}{{a}^{p}}=p$ we will have ${{\log }_{10}}\left( \dfrac{1}{10} \right)={{\log }_{10}}{{10}^{-1}}=-1$ .
Now we have $ {{\log }_{10}}\left( \dfrac{10x}{3} \right)-2=-1$ . Now by shifting $-2$ from left hand side to right hand side we will have
$\begin{align}
  & \Rightarrow {{\log }_{10}}\left( \dfrac{10x}{3} \right)-2=-1 \\
 & \Rightarrow {{\log }_{10}}\left( \dfrac{10x}{3} \right)=2-1=1 \\
\end{align}$ .
Now we can say that
 $\begin{align}
  & \Rightarrow {{\log }_{10}}\left( \dfrac{10x}{3} \right)=1 \\
 & \Rightarrow {{\log }_{10}}\left( \dfrac{10x}{3} \right)={{\log }_{10}}\left( 10 \right) \\
\end{align}$ .
Since ${{\log }_{10}}10=1$ from ${{\log }_{a}}{{a}^{p}}=p$ .
Hence we can say that $\dfrac{10x}{3}=10$

Therefore we can conclude that the value of $x$ in the given logarithmic expression ${{\log }_{10}}\left( \dfrac{10x}{3} \right)-2={{\log }_{10}}\left( \dfrac{1}{10} \right)$ is given as $x=3$.

Note: While answering questions of this type we should be sure with our logarithmic concepts that we are going to apply in the process and the calculations that we are going to perform in between the steps. This is a very simple and easy question and can be answered accurately in a short span of time. Very few mistakes are possible in questions of this type. Similarly we have many other logarithmic identities for example consider $\log \left( \dfrac{a}{b} \right)=\log a-\log b$.