Show that f(x) = log_{a} x, 0 < a < 1 is a decreasing function for all x > 0.

Given:- Function f(x) = log_{a} x , 0 < a < 1

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = log_{a} x, 0 < a < 1

⇒

⇒

As given 0 < a < 1

⇒ log(a) < 0

and for x > 0

⇒

Therefore f’(x) is

⇒

⇒ f’(x) < 0

Hence, condition for f(x) to be decreasing

Thus f(x) is decreasing for all x > 0

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