If int_{log _{e} 2}^{x} (e^{t}-1)^{-1} dt = l | vedclass

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(C) Given the integral $\int_{\log _{e} 2}^{x} \frac{1}{e^{t}-1} dt = \log _{e} \frac{3}{2}$.Let $u = e^{t}-1$, then $du = e^{t} dt$, which implies $d

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$\log _{e} 4$

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(C) Given the integral $\int_{\log _{e} 2}^{x} \frac{1}{e^{t}-1} dt = \log _{e} \frac{3}{2}$.Let $u = e^{t}-1$, then $du = e^{t} dt$, which implies $dt = \frac{du}{u+1}$.Substituting this into the integral, we get $\int \frac{1}{u(u+1)} du = \int (\frac{1}{u} - \frac{1}{u+1}) du = \log _{e} |u| - \log _{e} |u+1| = \log _{e} |\frac{u}{u+1}|$.Substituting $u = e^{t}-1$, the integral becomes $[\log _{e} |\frac{e^{t}-1}{e^{t}}|]_{\log _{e} 2}^{x} = [\log _{e} |1-e^{-t}|]_{\log _{e} 2}^{x}$.Evaluating the limits: $\log _{e} (1-e^{-x}) - \log _{e} (1-e^{-\log _{e} 2}) = \log _{e} (1-e^{-x}) - \log _{e} (1-\frac{1}{2}) = \log _{e} (1-e^{-x}) - \log _{e} (\frac{1}{2}) = \log _{e} \frac{3}{2}$.Thus, $\log _{e} (\frac{1-e^{-x}}{1/2}) = \log _{e} \frac{3}{2}$, which implies $2(1-e^{-x}) = \frac{3}{2}$.$1-e^{-x} = \frac{3}{4} \implies e^{-x} = 1 - \frac{3}{4} = \frac{1}{4}$.Taking the natural logarithm on both sides, $-x = \log _{e} (\frac{1}{4}) = -\log _{e} 4$.Therefore, $x = \log _{e} 4$.

If $\int_{\log _{e} 2}^{x} (e^{t}-1)^{-1} dt = \log _{e} \frac{3}{2}$, then the value of $x$ is

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