Find x, if: $x-\log 48+3\log 2=\dfrac{1}{3}\log 125-\log 3$.

Hint: In this question, we are given an equation involving x and logarithmic functions. We need to find the value of x. For this, we will first arrange the equation such that x is on one side and logarithmic functions are on the other side. After that, we will use various properties of logarithmic to evaluate our answer. Different logarithmic properties that we will use are:
\[\begin{align}
  & \left( i \right)\log {{m}^{n}}=n\log m \\
 & \left( ii \right)\log m+\log n=\log \left( m\times n \right) \\
 & \left( iii \right)\log m-\log n=\log \left( \dfrac{m}{n} \right) \\
 & \left( iv \right)\log 10=1 \\
\end{align}\]

Complete step by step answer:
Here we are given the equation as:
\[x-\log 48+3\log 2=\dfrac{1}{3}\log 125-\log 3\]
Let us rearrange this equation such that x remains on the left side and all logarithmic terms go towards the right side of the equation. Hence, we get:
\[x=\dfrac{1}{3}\log 125-\log 3+\log 48-3\log 2\cdots \cdots \cdots \cdots \left( 1 \right)\]
Now, let us simplify $\dfrac{1}{3}\log 125\text{ and }3\log 2$ separately. Simplifying $\dfrac{1}{3}\log 125$.
As we know, $\log {{m}^{n}}=n\log m$ so we can apply it here and we get $\log {{\left( 125 \right)}^{\dfrac{1}{3}}}$.
Now 125 can he written as $5\times 5\times 5$ so $\log {{\left( 125 \right)}^{\dfrac{1}{3}}}$ becomes $\log {{\left( 5\times 5\times 5 \right)}^{\dfrac{1}{3}}}\Rightarrow \log {{\left( {{5}^{3}} \right)}^{\dfrac{1}{3}}}$
As we know, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ so we get $\log {{5}^{\dfrac{3}{3}}}=\log 5$ so,
\[\dfrac{1}{3}\log 125=\log 5\cdots \cdots \cdots \left( 2 \right)\]
Simplifying $3\log 2$ as we know, $\log {{m}^{n}}=n\log m$ so we can apply it here and we get $\log {{\left( 2 \right)}^{3}}$.
Since ${{2}^{3}}=2\times 2\times 2=8$ so we get $\log 8$.
Therefore, \[3\log 2=\log 8\cdots \cdots \cdots \left( 3 \right)\]
Now putting values of (2) and (3) in (1) we get:
\[\Rightarrow x=\log 5-\log 3+\log 48-\log 8\]
Taking negative sign common from log 3 and log 8 and arranging the equation, we get:
\[\Rightarrow x=\log 5+\log 48-\left( \log 3+\log 8 \right)\]
We know, logarithmic property given as $\log m+\log n=\log \left( m\times n \right)$ so applying it on $\left( \log 5+\log 48 \right)\text{ and }\left( \log 3+\log 8 \right)$ separately, we get:
\[\begin{align}
  & \Rightarrow x=\log \left( 5\times 48 \right)-\log \left( 3\times 8 \right) \\
 & \Rightarrow x=\log 240-\log 24 \\
\end{align}\]
We know, logarithmic property given as $\log m-\log n=\log \left( \dfrac{m}{n} \right)$ so applying it on above equation, we get:
\[\begin{align}
  & \Rightarrow x=\log \left( \dfrac{240}{24} \right) \\
 & \Rightarrow x=\log 10 \\
\end{align}\]
As we know, the value of log10 is equal to 1, so the value of x becomes equal to 1.

Therefore, x = 1.

Note: Students should keep in mind all the logarithmic properties and carefully apply them. In this question, no base is given but we know log word is used for base 10 that is why log 10 is taken as 1. Take care of signs while applying logarithmic properties. Keep in mind that there is no property of $\log m\times \log n=\log \left( m+n \right),\dfrac{\log m}{\log n}=\log \left( m-n \right)$. Students can get confused in these properties.