Question

The value of $\left({\log }_{3}11\right)\left({\log }_{11}13\right)\left({\log }_{13}15\right)\left({\log }_{15}27\right)\left({\log }_{27}81\right)$ is equal to

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

Answer: (C)

$4$

let $a=\left({\log }_{3}11\right)\left({\log }_{11}13\right)\left({\log }_{13}15\right)\left({\log }_{15}27\right)\left({\log }_{27}81\right)$

$\Rightarrow a=\left(\frac{\log 11}{\log 3}\right)\left(\frac{\log 13}{\log 11}\right)\left(\frac{\log 15}{\log 13}\right)\left(\frac{\log 27}{\log 15}\right)\left(\frac{\log 81}{\log 27}\right)[\because {\log }_{b}a=\frac{\log a}{\log b}]$

$\Rightarrow a=\frac{\log 81}{\log 3}=\frac{\log 3^4}{\log 3}$

$\Rightarrow a=\frac{4\log 3}{\log 3}[\because \log a^m=m\log a]$

$\Rightarrow a=4$
Ans: C