Question
Write the set of values of $'a'$ for which $\text{f}(\text{x})=\log_\text{a}\text{x}$ is increasing in its domain.

Answer

$\text{f}(\text{x})=\log_\text{a}\text{x}$ Let, $\text{x}_1,\text{x}_2\in(0,\infty)$ such that $x_1< x_2.$ Since given function is logorithmic, either $a > 1$ or $0 < a < 1.$
Case I:
Let $a > 1$ Here, $\text{x}_1<\text{x}_2$
$\Rightarrow\log_\text{a}\text{x}_1<\log_\text{a}\text{x}_2$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$
$\therefore\ \text{x}_1< \text{x}_2$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$
Thus, for $a > 1, f(x)$ is increasing on $(0,\infty).$
Case II:
Let $0 < a < 1$ Here, $\text{x}_1<\text{x}_2$
$\Rightarrow\log_\text{a}\text{x}_1>\log_\text{a}\text{x}_2$
$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$
$\therefore\ \text{x}_1< \text{x}_2$
$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$
​​​​​​​Thus, for $a > 1, f(x)$ is increasing in its domain.

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