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

Question

Dr. Abubacker · Accepted Answer

(b)
$Given.\ \ area = \int_{0}^{2}{2^{kx}\ \ dx = \frac{3}{\log{2\ }}}$
$\Longrightarrow \ \ \left\lbrack \frac{2^{kx}}{\log_{e}2} \right\rbrack_{0}^{2} = \frac{3}{\log 2}$
$\Longrightarrow \ \ \frac{2^{2k}}{\log_{e}2} - \frac{1}{\log_{e}2} = \frac{3}{\log 2}$
$\Longrightarrow \ \ 2^{2k} - 1 = 3$
$\Longrightarrow \ 2^{2k} = 2^{2}$
$\Longrightarrow 2k = 2$
$\Longrightarrow k = 1$

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

Explanation

Reviewed by Dr. Abubacker

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