Question

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

Answer 1

Find the value of the given expression

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

The value of the given expression is

$\begin{array}{rcl}{\log }_3\left(4\right)·{\log }_4\left(5\right)·{\log }_5\left(6\right)·{\log }_6\left(7\right)·{\log }_7\left(8\right)·{\log }_8\left(9\right)& =& \frac{{\log }_{10}\left(4\right)}{{\log }_{10}\left(3\right)}·\frac{{\log }_{10}\left(5\right)}{{\log }_{10}\left(4\right)}·\frac{{\log }_{10}\left(6\right)}{{\log }_{10}\left(5\right)}·\frac{{\log }_{10}\left(7\right)}{{\log }_{10}\left(6\right)}·\frac{{\log }_{10}\left(8\right)}{{\log }_{10}\left(7\right)}·\frac{{\log }_{10}\left(9\right)}{{\log }_{10}\left(8\right)}\\ & =& \frac{{\log }_{10}\left(9\right)}{{\log }_{10}\left(3\right)}\\ & =& \frac{{\log }_{10}{\left(3\right)}^2}{{\log }_{10}\left(3\right)}\\ & =& 2\end{array}$ $\left[\because {\log }_a\left(b\right)=\frac{{\log }_{10}\left(b\right)}{{\log }_{10}\left(a\right)}\right]$

Hence, the required answer is $2$.