If $\frac{1}{\log_3 \pi} + \frac{1}{\log_4 \pi} > x$,then $x =$ ?

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 $x$ satisfying $\log _{1/2}(x^2 - 6x + 12) \ge -2$ is

The number of solutions of the equation $\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}-4=0$ for $x > 0$ is:

Let $a = 3 \sqrt{2}$ and $b = \frac{1}{5^{\frac{1}{6}} \sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that $3x + 2y = \log_a(18)^{\frac{5}{4}}$ and $2x - y = \log_b(\sqrt{1080})$,then $4x + 5y$ is equal to:

For $y = \log_a x$ to be defined,$a$ must be: