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(B) The area $A$ is given by the integral $\int_{0}^{2} 2^{kx} \, dx = \frac{3}{\ln 2}$.Evaluating the integral: $\int_{0}^{2} 2^{kx} \, dx = \left[ \

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$1$

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(B) The area $A$ is given by the integral $\int_{0}^{2} 2^{kx} \, dx = \frac{3}{\ln 2}$.Evaluating the integral: $\int_{0}^{2} 2^{kx} \, dx = \left[ \frac{2^{kx}}{k \ln 2} \right]_{0}^{2}$.Substituting the limits: $\frac{1}{k \ln 2} (2^{2k} - 2^0) = \frac{2^{2k} - 1}{k \ln 2}$.Equating this to the given area: $\frac{2^{2k} - 1}{k \ln 2} = \frac{3}{\ln 2}$.This simplifies to $\frac{2^{2k} - 1}{k} = 3$,or $2^{2k} - 1 = 3k$.Testing the options:For $k = 1$: $2^{2(1)} - 1 = 4 - 1 = 3$,and $3(1) = 3$. Since $3 = 3$,the condition is satisfied.Thus,the value of $k$ is $1$.

If the area above the $x$-axis,bounded by the curves $y = 2^{kx}$,$x = 0$,and $x = 2$ is $\frac{3}{\ln 2}$,then the value of $k$ is

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