Prove the equation ${{\log }_{10}}125=3-3{{\log }_{10}}2$

Hint: To solve this question, we should know the properties of logarithms. For given numbers of a, b, c…we can write ${{\log }_{x}}\left( \dfrac{abc...}{pqr..} \right)={{\log }_{x}}a+{{\log }_{x}}b+{{\log }_{x}}c...-{{\log }_{x}}p-{{\log }_{x}}q-{{\log }_{x}}r...$. We can write the factorisation of 125 and after that, we can write the number 5 as $\dfrac{10}{2}$ to get the required answer.

Complete step-by-step solution:
Let us consider the term ${{\log }_{10}}125$. To solve further, we should do the factorisation of 125. By doing factorisations, we get
$\begin{align}
  & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
From the above factorisation, we can write that
$125=5\times 5\times 5$
Using this value of 125 in the logarithm, we get
${{\log }_{10}}125={{\log }_{10}}5\times 5\times 5$
We know the formula of logarithms which is
For given numbers of a, b, c…we can write ${{\log }_{x}}\left( \dfrac{abc...}{pqr..} \right)={{\log }_{x}}a+{{\log }_{x}}b+{{\log }_{x}}c...-{{\log }_{x}}p-{{\log }_{x}}q-{{\log }_{x}}r...\to \left( 1 \right)$
Using this formula, we can write the value of ${{\log }_{10}}125$ as
${{\log }_{10}}125={{\log }_{10}}5+{{\log }_{10}}5+{{\log }_{10}}5=3{{\log }_{10}}5\to \left( 2 \right)$
We have to get the R.H.S into the form of $3-3{{\log }_{10}}2$, which means that we should include 2 into the logarithms. We have the logarithmic properties in terms of multiplication and divisions. So, we have to get a relation between 5 and 2 in terms of multiplication. We can write the multiplicative relation between 5 and 2 as
$5=\dfrac{10}{2}$
Using this in equation-2, we get
$3{{\log }_{10}}5=3{{\log }_{10}}\left( \dfrac{10}{2} \right)$
Using the equation-1, we get
$3{{\log }_{10}}5=3{{\log }_{10}}\left( \dfrac{10}{2} \right)=3\left( {{\log }_{10}}10-{{\log }_{10}}2 \right)$
We know that for any given positive number a, ${{\log }_{a}}a=1$
Using this relation, we get
$\begin{align}
  & 3{{\log }_{10}}5=3{{\log }_{10}}\left( \dfrac{10}{2} \right)=3\left( 1-{{\log }_{10}}2 \right)=3-3{{\log }_{10}}2 \\
 & {{\log }_{10}}125=3-3{{\log }_{10}}2 \\
\end{align}$
$\therefore $ Hence proved the equation ${{\log }_{10}}125=3-3{{\log }_{10}}2$.

Note: An alternate approach to prove the given statement is by considering ${{\log }_{10}}125+3{{\log }_{10}}2$. We can write the considered expression as${{\log }_{10}}125+{{\log }_{10}}2+{{\log }_{10}}2+{{\log }_{10}}2$. We know the formula related to logarithms and by applying it, we get ${{\log }_{10}}125+3{{\log }_{10}}2={{\log }_{10}}\left( 125\times 2\times 2\times 2 \right)={{\log }_{10}}1000={{\log }_{10}}\left( 10\times 10\times 10 \right)$.
We can write ${{\log }_{10}}\left( 10\times 10\times 10 \right)={{\log }_{10}}10+{{\log }_{10}}10+{{\log }_{10}}10=3{{\log }_{10}}10=3$