The solution of the equation ${\log _7}{\log _5}$ $(\sqrt {{x^2} + 5 + x} ) = 0$
$x = 2$
$x = 3$
$x = 4$
$x = - 2$
If ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0,$ then $x$ is equal to
If ${x^{{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}}} = \sqrt 3 $ then $x$ has
If $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2,$ then $A$ is equal to
The value of $6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \ldots}}}\right)$ is
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 $\frac{x}{y}$