If ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0,$ then $x$ is equal to
$k$
${1 \over 5}$
$5$
$(b)$ and $(c)$ both
The set of real values of $x$ for which ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ is
If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$ then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}{x_n}^{x_{n - 1}^{{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} ^{{x_1}}}}}$ is equal to
If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1),$ then $x$ lies in the interval
Let $x, y$ be real numbers such that $x>2 y>0$ and $2 \log (x-2 y)=\log x+\log y$ Then, the possible value(s) of $\frac{x}{y}$
If ${\log _{10}}2 = 0.30103,{\log _{10}}3 = 0.47712,$ the number of digits in ${3^{12}} \times {2^8} $ is