Let f(x) be defined for all x 0 and be cont | vedclass

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(D) Given the functional equation $f(\frac{x}{y}) = f(x) - f(y)$.Setting $y = 1$,we get $f(x) = f(x) - f(1)$,which implies $f(1) = 0$.Setting $x = 1$,

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$f(x) = \ln(x)$

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(D) Given the functional equation $f(\frac{x}{y}) = f(x) - f(y)$.Setting $y = 1$,we get $f(x) = f(x) - f(1)$,which implies $f(1) = 0$.Setting $x = 1$,we get $f(\frac{1}{y}) = f(1) - f(y) = -f(y)$.For any $x, y > 0$,$f(xy) = f(\frac{x}{1/y}) = f(x) - f(1/y) = f(x) + f(y)$.This is the Cauchy functional equation for the logarithm function.Since $f(x)$ is continuous and satisfies $f(xy) = f(x) + f(y)$,the solution is of the form $f(x) = c \ln(x)$.Given $f(e) = 1$,we have $c \ln(e) = 1$,so $c(1) = 1$,which means $c = 1$.Thus,$f(x) = \ln(x)$.Therefore,option $D$ is correct.

Let $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f(\frac{x}{y}) = f(x) - f(y)$ for all $x, y > 0$ and $f(e) = 1$. Then:

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