If ${\log _{10}}x = y$,then the value of ${\log _{1000}}{x^2}$ is

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}}} = \sqrt{3}$,then $x$ has:

The number of solutions of the equation $\log_{\sqrt{3}}(x^{3} - 1) = \log_{\sqrt{3}}(x - 1) + 2$ is:

Given that $a, b \in \{0, 1, 2, \ldots, 9\}$ with $a+b \neq 0$ and that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

The set of real values of $x$ for which $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$ is

If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$,then $x$ lies in which interval?