A) \[L(x+y)=L(x)+L(y)\]
B) \[L\left( \frac{x}{y} \right)=L(x)+L(y)\]
C) \[L(xy)=L(x)+L(y)\]
D) None of these
Correct Answer: C
Solution :
Given function \[L(x)=\int_{1}^{x}{\frac{1}{t}dt=[\log t]_{1}^{x}}=\log x-\log 1\] Þ \[L(x)=\log x\], Hence \[L\,(xy)=L(x)+L(y)\].
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