Prove that the logarithmic function $f(x) = \log x$ is strictly increasing on $(0, \infty).$
(N/A) The given function is $f(x) = \log x$.
To determine the interval of increase,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$.
For the function to be strictly increasing,we require $f'(x) > 0$.
In the given interval $(0, \infty)$,$x$ is always positive $(x > 0)$.
Therefore,$\frac{1}{x} > 0$ for all $x \in (0, \infty)$.
Since $f'(x) > 0$ for all $x$ in the interval $(0, \infty)$,the function $f(x) = \log x$ is strictly increasing on $(0, \infty)$.
To determine the interval of increase,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$.
For the function to be strictly increasing,we require $f'(x) > 0$.
In the given interval $(0, \infty)$,$x$ is always positive $(x > 0)$.
Therefore,$\frac{1}{x} > 0$ for all $x \in (0, \infty)$.
Since $f'(x) > 0$ for all $x$ in the interval $(0, \infty)$,the function $f(x) = \log x$ is strictly increasing on $(0, \infty)$.
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