For $y = {\log _a}x$ to be defined $'a'$ must be
Any positive real number
Any number
$ \ge e$
Any positive real number $ \ne 1$
(d) It is obvious.
Let $a, b, x$ be positive real numbers with $a \neq 1$, $x \neq 1$, ab $\neq 1$. Suppose $\log _{ a } b =10$, and $\frac{\log _{ a } x \log _{ x }\left(\frac{ b }{ a }\right)}{\log _{ x } b \log _{ ab } x }=\frac{ p }{ q }$, where $p$ and $q$ are positive integers which are coprime. Then $p+q$ is
${\log _4}18$ is
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
Solution set of inequality ${\log _{10}}({x^2} – 2x – 2) \le 0$ is
If ${\log _{0.04}}(x – 1) \ge {\log _{0.2}}(x – 1)$ then $x$ belongs to the interval
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