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 ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$ then $x \ne 1$ lies in
The interval of $x$ in which the inequality ${5^{(1/4)(\log _5^2x)}}\, \geqslant \,5{x^{(1/5)(\log _5^x)}}$
If ${\log _5}a.{\log _a}x = 2,$then $x$ is equal to
${\log _4}18$ is
$7\log \left( {{{16} \over {15}}} \right) + 5\log \left( {{{25} \over {24}}} \right) + 3\log \left( {{{81} \over {80}}} \right)$ is equal to