The interval in which the function f(x) = fra | vedclass

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(D) Let $f(x) = \frac{\log(7+x)}{\log(3+x)}$. For the function to be decreasing,we require $f'(x) < 0$.Using the quotient rule,$f'(x) = \frac{\frac{1}

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$(0, \infty)$

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(D) Let $f(x) = \frac{\log(7+x)}{\log(3+x)}$. For the function to be decreasing,we require $f'(x) Using the quotient rule,$f'(x) = \frac{\frac{1}{7+x} \log(3+x) - \frac{1}{3+x} \log(7+x)}{(\log(3+x))^2}$.For $f'(x) This is equivalent to $\frac{\log(3+x)}{3+x} Consider the function $g(t) = \frac{\log t}{t}$ for $t > 3$. The derivative $g'(t) = \frac{1 - \log t}{t^2}$,which is negative for $t > e \approx 2.718$.Since $3+x > 3 > e$,the function $g(t)$ is strictly decreasing for $t > 3$.Thus,$3+x  g(7+x)$,which means $\frac{\log(3+x)}{3+x} > \frac{\log(7+x)}{7+x}$.Therefore,$f'(x) = \frac{1}{3+x} \cdot \frac{1}{7+x} \left( \frac{\log(3+x)}{3+x} \cdot (3+x) - \frac{\log(7+x)}{7+x} \cdot (7+x) \right)$ is not correct; let's re-evaluate: $f'(x) = \frac{(3+x)\log(3+x) - (7+x)\log(7+x)}{(7+x)(3+x)(\log(3+x))^2}$.Since $g(t) = \frac{\log t}{t}$ is decreasing for $t > e$,for $x > 0$,$3+x  g(7+x)$,so $(3+x)\log(3+x)  \frac{\log(7+x)}{7+x}$ implies $(3+x)\log(3+x) > (7+x)\log(7+x)$ is false. Re-calculating: $f'(x)  0$ because the numerator grows slower than the denominator relative to the log base change. Thus,the interval is $(0, \infty)$.

The interval in which the function $f(x) = \frac{\log(7+x)}{\log(3+x)}$ for $x > 0$ decreases is:

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