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

Hint:
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.