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The function $L(x)=\int_1^x \frac{d t}{t}$ satisfies the equation
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Verified Answer
The correct answer is:
$L(x y)=L(x)+L(y)$
Given function $L(x)-\int_1^x \frac{1}{t} d t=[\log t]_1^x=\log x-\log 1$
$\Rightarrow L(x)-\log x$, Hence $L(x y)=L(x)+L(y)$.
$\Rightarrow L(x)-\log x$, Hence $L(x y)=L(x)+L(y)$.
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