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If $\int_{\log _{e} 2}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e} \frac{3}{2}$ then the value of $x$ is
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The correct answer is:
$\log 4$
$\int_{\log _{e}{ }^{2}}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e}^{\frac{3}{2}} \int_{\log _{e}{ }^{2}}^{x} \frac{e^{-x}}{1-e^{-x}} d x=\log _{e}^{\frac{3}{2}}$
$e^{-x}=\frac{1}{4}$
$x=\ln 4$
$e^{-x}=\frac{1}{4}$
$x=\ln 4$
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