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 $\frac{x}{y}$

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

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

If ${\log _5}a.{\log _a}x = 2,$then $x$ is equal to

Let $a , b , c$ be three distinct positive real numbers such that $(2 a)^{\log _{\varepsilon} a}=(b c)^{\log _e b}$ and $b^{\log _e 2}=a^{\log _e c}$. Then $6 a+5 b c$ is equal to $........$.

If ${a^2} + 4{b^2} = 12ab,$ then $\log (a + 2b)$ is