Question

Prove that ${\log }_{3}2\times {\log }_{4}3\times {\log }_{5}4\times {\log }_{6}5\times {\log }_{7}6\times {\log }_{8}7=\frac{1}{3}$

Answer 1

${\log }_{3}2\times {\log }_{4}3\times {\log }_{5}4\times {\log }_{6}5\times {\log }_{7}6\times {\log }_{8}7$
$=({\log }_{3}2\times {\log }_{4}3)\times ({\log }_{5}4\times {\log }_{6}5)\times ({\log }_{7}6\times {\log }_{8}7)$
$={\log }_{4}2\times {\log }_{6}4\times {\log }_{8}6=({\log }_{4}2\times {\log }_{6}4)\times {\log }_{8}6$ $[\because {\log }_{a}M={\log }_{b}M\times {\log }_{a}b]$
$={\log }_{6}2\times {\log }_{8}6={\log }_{8}2=\frac{1}{{\log }_{2}8}$ $[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$=\frac{1}{{\log }_2{2}^3}=\frac{1}{3{\log }_{2}2}$ $[\because {\log }_a{\left(M\right)}^n=n{\log }_{a}M]$
$=\frac{1}{3}$ $[\because {\log }_{2}2=1]$