Question

The number $\text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( {{\log }_{6}}2 \right)}^{2}}$when simplified reduces to
A. A prime number
B. Irrational number
C. A real which is less than \[{{\log }_{3}}\pi \]
D. A real which is greater than\[{{\log }_{7}}6\]

Answer 1

Hint: Type of question is based on the simple log calculation. As we had given an equation which we need to simplify and relate the answer with the given option. So we will solve the given problem to reduce it to a simpler value using the log identity.

Complete step by step answer:
So moving ahead with the question, we have$\text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( {{\log }_{6}}2 \right)}^{2}}$
So by using the identity${{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$ in${{\left( {{\log }_{6}}2 \right)}^{2}}$, so we will get the new equation;
\[\text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( \dfrac{1}{{{\log }_{2}}6} \right)}^{2}}\]
Now by using the identity of ${{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n$in \[{{\log }_{2}}6\] so we will get the new equation;
\[\begin{align}
  & \text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( \dfrac{1}{{{\log }_{2}}2+{{\log }_{2}}3} \right)}^{2}} \\
 & \text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( \dfrac{1}{1+{{\log }_{2}}3} \right)}^{2}} \\
\end{align}\]
By using the identity ${{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$again. New equation we will get;
\[\begin{align}
  & \text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( \dfrac{1}{1+{{\log }_{2}}3} \right)}^{2}} \\
 & \text{N}=\dfrac{1+2{{\log }_{3}}2}{{{\left( 1+{{\log }_{3}}2 \right)}^{2}}}+{{\left( \dfrac{1}{1+\dfrac{1}{{{\log }_{3}}2}} \right)}^{2}} \\
\end{align}\]
Since, \[{{\log }_{3}}2\]is in the function, let us assume it as ‘a’ as it will make the equation seem to be simple. So new equation we will get;
\[\begin{align}
  & \text{N}=\dfrac{1+2a}{{{\left( 1+a \right)}^{2}}}+{{\left( \dfrac{1}{1+\dfrac{1}{a}} \right)}^{2}} \\
 & \text{N}=\dfrac{1+2a}{{{\left( 1+a \right)}^{2}}}+{{\left( \dfrac{a}{a+1} \right)}^{2}} \\
 & \text{N}=\dfrac{1+2a}{{{\left( 1+a \right)}^{2}}}+{{\left( \dfrac{a}{a+1} \right)}^{2}} \\
\end{align}\]
Since both have same denominator seo we can directly add the equation, so we will get the equation;
 \[\begin{align}
  & \text{N}=\dfrac{1+2a}{{{\left( 1+a \right)}^{2}}}+{{\left( \dfrac{a}{a+1} \right)}^{2}} \\
 & \text{N}=\dfrac{1+2a+{{a}^{2}}}{{{\left( 1+a \right)}^{2}}} \\
\end{align}\]
Since we know that \[{{\left( 1+a \right)}^{2}}=1+2a+{{a}^{2}}\]is an identity, so we can write the above equation as;
\[\begin{align}
  & \text{N}=\dfrac{1+2a+{{a}^{2}}}{{{\left( 1+a \right)}^{2}}} \\
 & \text{N}=\dfrac{{{\left( 1+a \right)}^{2}}}{{{\left( 1+a \right)}^{2}}} \\
 & N=1 \\
\end{align}\]
Hence all values cancel and we get the answer 1.
Since 1 is not a prime number. And not an irrational number. As the c option says that its value should be less than \[{{\log }_{3}}\pi \]which is approx. greater than 1, hence we can say this option is correct.

So, the correct answer is “Option C”.

Note: As we had assumed the \[{{\log }_{3}}2\]value a variable ’a’ which ultimately get cancel at last, so answer we got is independent of \[{{\log }_{3}}2\]. Moreover if you do it without assuming the value with some variable then also you will get the same answer, by assuming we have only motive to have a simple look of the equation.