The interval in which the function \(f(x)=\frac{\log (7+x)}{\log (3+x)}(x>0)\) decreases is
The interval in which the function \(f(x)=\frac{\log (7+x)}{\log (3+x)}(x>0)\) decreases is
\(\left(0, \frac{7}{3}\right)\)
\(\left(0, \frac{3}{7}\right)\)
\((0,1)\)
\((0, \infty)\)
Solution
If the given function \(f(x)=\frac{\log (7+x)}{\log (3+x)}, x > 0\) is decreasing function then \(f^{\prime}(x) < 0\)
\(\Rightarrow \quad \frac{\log (3+x) \frac{1}{(7+x)}-\frac{1}{(3+x)} \log (7+x)}{[\log (3+x)]^2} < 0\)
\(\Rightarrow \quad(3+x)^{(3+x)} < (7+x)^{(7+x)}\), it is true for every value of \(x > 0\).
\(\Rightarrow \quad x \in(0, \infty)\)
Hence, option (d) is correct.