(B) Step $1$: Simplify the integrand using the property $e^{\log a} = a$. We have $e^{-x \log 2} = (e^{\log 2})^{-x} = 2^{-x}$.Step $2$: Substitute th

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(B) Step $1$: Simplify the integrand using the property $e^{\log a} = a$. We have $e^{-x \log 2} = (e^{\log 2})^{-x} = 2^{-x}$.Step $2$: Substitute this into the integral: $\int 2^{-x} \cdot 2^x dx$.Step $3$: Simplify the product: $\int 2^{-x+x} dx = \int 2^0 dx = \int 1 dx$.Step $4$: Integrate with respect to $x$: $\int 1 dx = x + C$.