The solution of the equation ${\log _7}{\log _5}$ $(\sqrt {{x^2} + 5 + x} ) = 0$
$x = 2$
$x = 3$
$x = 4$
$x = - 2$
If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$ then $x$ belongs to the interval
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
If ${\log _{10}}x = y,$ then ${\log _{1000}}{x^2} $ is equal to
If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$ then $x \ne 1$ lies in
The value of ${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}$ is