Question

The function \[L\left( x \right) = \int\limits_1^x {\dfrac{{dt}}{t}} \] satisfies which of the following equation?
A. \[L\left( {x + y} \right) = L\left( x \right) + L\left( y \right)\]
B. \[L\left( {\dfrac{x}{y}} \right) = L\left( x \right) + L\left( y \right)\]
C. \[L\left( {xy} \right) = L\left( x \right) + L\left( y \right)\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, solve the integral by using the formula \[\int {\dfrac{1}{x}} dx = \log x\]. Then, further solve it by applying the limit and calculate the value of \[L\left( x \right)\].
After that, check which of the given equation satisfy the value of \[L\left( x \right)\].



Formula Used:Integration Formula: \[\int {\dfrac{1}{x}} dx = \log x\]
The product property of the logarithm: \[\log \left( {ab} \right) = \log a + \log b\]



Complete step by step solution:The given function is \[L\left( x \right) = \int\limits_1^x {\dfrac{{dt}}{t}} \].
Let solve the right-hand side of the given function by applying the integration formula \[\int {\dfrac{1}{x}} dx = \log x\].
We get,
\[L\left( x \right) = \left[ {\log t} \right]_1^x\]
Apply the upper and lower limits.
\[ \Rightarrow L\left( x \right) = \log x - \log 1\]
\[ \Rightarrow L\left( x \right) = \log x - 0\]
\[ \Rightarrow L\left( x \right) = \log x\]

We know the product property of the logarithmic function \[\log \left( {ab} \right) = \log a + \log b\].
By applying this product property of the logarithm on the above function, we get
\[L\left( {xy} \right) = L\left( x \right) + L\left( y \right)\]



Option ‘C’ is correct



Note: Students often get confused about the properties of the logarithmic function. They think \[\log \left( {a + b} \right) = \log a + \log b\] is a property of logarithm. But it is wrong. The correct property is \[\log \left( {ab} \right) = \log a + \log b\].