Question

Prove that the function $f(x) = {\log _a}x$ is increasing on $(0,\infty )$ , if $a > 1$ and decreasing on $(0,\infty )$ , if $0 < a < 1$ .

Answer 1

A function $f(x) = {\log _a}x$ is given to us in this case. If differentiating it with respect to x gives us a positive result we have an increasing function and if it is negative we have it as a decreasing function. So we use this concept and we apply different conditions on a and prove the given statement.

Complete step by step solution:
We are given a function, $f(x) = {\log _a}x$
It can be written as, $f\left( x \right) = \dfrac{{{{\log }_e}x}}{{{{\log }_e}a}}$
Now, differentiating with respect to x, we get,
Since $\dfrac{{d{{\log }_e}x}}{{dx}} = \dfrac{1}{x}$ ,
\Rightarrow $$$$f'\left( x \right) = \dfrac{1}{{{{\log }_e}a}} \times \dfrac{1}{x}
On simplification we get,
\Rightarrow $$$$f'\left( x \right) = \dfrac{1}{{x{{\log }_e}a}}
Now, if $a > 1$ ,
Then we get,
$f\prime \left( x \right) > 0$
$\therefore f\left( x \right)\;$ is increasing function for $a > 1$
$f'\left( x \right) = \dfrac{1}{{x{{\log }_e}a}}$
if $a > 0\;\& \;a < 1$
Then $f\prime \left( x \right)\;$ is not greater than zero.
i.e., $f'(x) < 0$

$\therefore f\left( x \right)\;$is decreasing function $0 < a < 1.$

Note:
Let $y = f(x)$ be a differentiable function on an interval $(a,b).\;$If for any two points ${x_1},{x_2} \in (a,b)\;$such that ${x_1} < {x_2}$, there holds the inequality $f({x_1}) \leqslant f({x_2}),\;$the function is called increasing (or non-decreasing ) in this interval.