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

Given:- Function f(x) = loga 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) = loga 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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