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Algebra of Derivatives

fx=logaxx>0 i...

Question

$f (x) = {log}_{a} x (x > 0)$ is an increasing function if

a > 1

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0 < a < 1

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a \neq 1

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a \neq 0

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Solution

The correct option is A $a > 1$
$f (x) = \frac{l o g x}{l o g a}$
$f^{'} (x) = \frac{1}{x l o g a}$
$f^{'} (x) > 0$ If $a > 1$

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f (x) = \{\begin{array}{l} \frac{1 - x^{n}}{1 - x}, & x \neq 1 \\ n - 1, & x = 1 \end{array} n \in N at x = 1

f (x) = \{\begin{cases} \frac{|x^{2} - 1|}{x - 1}, for & x \neq 1 \\ 2, for & x = 1 \end{cases} at x = 1

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