If ${\log _4}5 = a$ and ${\log _5}6 = b,$ then ${\log _3}2$ is equal to
${1 \over {2a + 1}}$
${1 \over {2b + 1}}$
$2ab + 1$
${1 \over {2ab - 1}}$
If ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0,$ then $x$ is equal to
The value of ${81^{(1/{{\log }_5}3)}} + {27^{{{\log }_{_9}}36}} + {3^{4/{{\log }_{_7}}9}}$ is equal to
The value of ${\log _3}\,4{\log _4}\,5{\log _5}\,6{\log _6}\,7{\log _7}\,8{\log _8}\,9$ is
The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :
Let $\log _a b=4, \log _c d=2$, where $a, b, c, d$ are natural numbers. Given that $b-d=7$, the value of $c-a$ is