Question

$\int e^{-x \log 2} 2^x dx =$

• A. $\log x + C$
• B. $x + C$
• C. $\frac{1}{x} + C$
• D. $\frac{x^2}{2} + C$

Hint:

Always simplify the integrand before trying complex integration techniques like integration by parts. The identity $e^{k \ln(a)} = a^k$ is a very common tool used by examiners to hide simple constants or terms within intimidating-looking exponential expressions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem involves an integral with an exponential term containing a logarithm in its exponent, multiplied by another exponential term. The key is to simplify the integrand using logarithmic and exponential properties before attempting integration.

Step 2: Key Formula or Approach:
Use the property of logarithms that $a \log b = \log(b^a)$ and the fundamental property relating exponentials and natural logarithms: $e^{\ln(y)} = y$. In higher-level mathematics, '$\log$' without a specified base usually implies the natural logarithm '$\ln$'.

Step 3: Detailed Explanation:
The given integral is: \[ I = \int e^{-x \log 2} \cdot 2^x dx \] Assuming $\log$ denotes the natural logarithm $\ln$: First, simplify the term $e^{-x \log 2}$. Using the power rule for logarithms ($-x \ln 2 = \ln(2^{-x})$): \[ e^{-x \ln 2} = e^{\ln(2^{-x})} \] Using the inverse property $e^{\ln(y)} = y$: \[ e^{\ln(2^{-x})} = 2^{-x} \] Now, substitute this simplified expression back into the integrand: \[ I = \int (2^{-x}) \cdot 2^x dx \] Use exponent rules to combine terms with the same base ($a^m \cdot a^n = a^{m+n}$): \[ I = \int 2^{-x + x} dx \] \[ I = \int 2^0 dx \] Since any non-zero number to the power of 0 is 1: \[ I = \int 1 dx \] The integral of a constant 1 with respect to $x$ is $x$: \[ I = x + C \] (Note: The handwritten tick mark on option 4 in the image is incorrect. The mathematically sound result based on the printed text is $x+C$.)

Step 4: Final Answer:
The integral evaluates to $x + C$.