Question

The value of $\frac{1+2{\log }_{3}2}{(1+{\log }_{3}2{)}^2}+({\log }_{6}2{)}^2$ is

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

The correct option is D 1

Let $y=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+{\left({\log }_{6}2\right)}^2$
$\Rightarrow y=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+{\left(\frac{{\log }_{3}2}{{\log }_{3}6}\right)}^2,[\because {\log }_{a}b=\frac{\log b}{\log a}]$
$\Rightarrow y=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+{\left(\frac{{\log }_{3}2}{{\log }_{3}3+{\log }_{3}2}\right)}^2,[\because \log (ab)=\log a+\log b\text{\&}{\log }_{a}a=1]$
$\Rightarrow y=\frac{1+2{\log }_{3}2{+\left({\log }_{3}2\right)}^2}{{(1+{\log }_{3}2)}^2}=\frac{{(1+{\log }_{3}2)}^2}{{(1+{\log }_{3}2)}^2}$
$\Rightarrow y=1$
Ans: D