If ∫ln⁡2x ex−1−1dx=ln⁡32 then x =

∫ln2x 1ex−1dx=ln⁡32⇒∫ln2x e−x1−e−xdx=ln⁡32⇒ln1-e-xln2x=ln32 ⇒ln1-e-x-ln12=ln32 ⇒ex-1ex=34 ⇒4ex-4=3ex ⇒ex=4⇒x=ln4

If ∫ln⁡2x ex−1−1dx=ln⁡32 then x =

∫ln2x 1ex−1dx=ln⁡32⇒∫ln2x e−x1−e−xdx=ln⁡32⇒ln1-e-xln2x=ln32 ⇒ln1-e-x-ln12=ln32 ⇒ex-1ex=34 ⇒4ex-4=3ex ⇒ex=4⇒x=ln4

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If $\int_{\ln 2}^{x} {(e^{x} - 1)}^{- 1} dx = \ln (\frac{3}{2})$ then x =

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$\ln 4$

$1$

$e^{2}$

$\frac{1}{e}$

answer is A.

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Detailed Solution

$\int_{\ln_{2}}^{x} \frac{1}{e^{x} - 1} dx = \ln (\frac{3}{2}) \Rightarrow \int_{\ln_{2}}^{x} \frac{e^{- x}}{1 - e^{- x}} dx = \ln (\frac{3}{2})$
$\Rightarrow {[\ln |1 - e^{- x}|]}_{\ln 2}^{x} = \ln (\frac{3}{2}) \Rightarrow \ln |1 - e^{- x}| - \ln (\frac{1}{2}) = \ln (\frac{3}{2}) \Rightarrow \frac{e^{x} - 1}{e^{x}} = \frac{3}{4} \Rightarrow 4 e^{x} - 4 = 3 e^{x} \Rightarrow e^{x} = 4 \Rightarrow x = \ln 4$