Question

The value of ${\log }_3\left(4\right){\log }_4\left(5\right){\log }_5\left(6\right){\log }_6\left(7\right){\log }_7\left(8\right){\log }_8\left(9\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

Explanation for the correct answer:

Compute the required value:

We know that ${\log }_b\left(a\right)=\frac{\log \left(a\right)}{\log \left(b\right)}$

Hence the given expression can be written as

${\log }_3\left(4\right){\log }_4\left(5\right){\log }_5\left(6\right){\log }_6\left(7\right){\log }_7\left(8\right){\log }_8\left(9\right)=\frac{\log \left(4\right)}{\log \left(3\right)}\times \frac{\log \left(5\right)}{\log \left(4\right)}\times \frac{\log \left(6\right)}{\log \left(5\right)}\times \frac{\log \left(7\right)}{\log \left(6\right)}\times \frac{\log \left(8\right)}{\log \left(7\right)}\times \frac{\log \left(9\right)}{\log \left(8\right)}$

$=\frac{\log \left(9\right)}{\log \left(3\right)}$

$=\frac{\log \left(3^2\right)}{\log \left(3\right)}$ $\left[\because \log \left(a^m\right)=m\log \left(a\right)\right]$

$=\frac{2\log \left(3\right)}{\log \left(3\right)}$

$\Rightarrow$ ${\log }_3\left(4\right){\log }_4\left(5\right){\log }_5\left(6\right){\log }_6\left(7\right){\log }_7\left(8\right){\log }_8\left(9\right)$$=2$

Hence the value of ${\log }_3\left(4\right){\log }_4\left(5\right){\log }_5\left(6\right){\log }_6\left(7\right){\log }_7\left(8\right){\log }_8\left(9\right)$ is $2$.

Hence, option $\left(B\right)$ is the correct answer.