Question

Prove that the function f(x) = log_a x is increasing on (0, ∞) if a > 1 and decreasing on (0, ∞), if 0 < a < 1.

Answer 1

$$\begin{gathered} f\left(x\right)={\log }_{a}x \\ \text{Let}{x}_1,x_2\in \left(0,\infty \right)\text{such that}{x}_1<x_2. \\ \text{Case 1: Let}a>1 \\ \text{Here,} \\ {x}_1<x_2 \\ \Rightarrow {\log }_a{x}_1<{\log }_a{x}_2 \\ \Rightarrow f\left(x_1\right)<f\left(x_2\right) \\ \therefore x_1<x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right),\forall x_1,x_2\in \left(0,\infty \right) \\ \text{So,}f\left(x\right)\text{is increasing on}\left(0,\infty \right)\text{.} \\ \text{Case 2: Let}0<a<1 \\ \text{Here,} \\ {x}_1<x_2 \\ \Rightarrow {\log }_a{x}_1>{\log }_a{x}_2 \\ \Rightarrow f\left(x_1\right)>f\left(x_2\right) \\ \therefore x_1<x_2\Rightarrow f\left(x_1\right)>f\left(x_2\right),\forall x_1,x_2\in \left(0,\infty \right) \\ \text{So,}f\left(x\right)\text{is decreasing on}\left(0,\infty \right)\text{.} \end{gathered}$$