Value of x satisfying ∫log⁡ 2x dxex−1=π6 is

Substituting ex−1=t2, exdx=2tdtdx=2tdtt2+1, so π6=∫log⁡ 2x dxex−1=∫1ex−1 2tdtt2+1t=2∫1ex−1 dtt2+1=2tan−1⁡ t1ex−1=2tan−1⁡ ex−1−π2⇒ 2tan−1⁡ ex−1=2π3⇒ ex−1=tan⁡ π3=3⇒ ex=1=3 i.e. ex=4⇒x=log⁡ 4=2log2

Value of x satisfying ∫log⁡ 2x dxex−1=π6 is

Substituting ex−1=t2, exdx=2tdtdx=2tdtt2+1, so π6=∫log⁡ 2x dxex−1=∫1ex−1 2tdtt2+1t=2∫1ex−1 dtt2+1=2tan−1⁡ t1ex−1=2tan−1⁡ ex−1−π2⇒ 2tan−1⁡ ex−1=2π3⇒ ex−1=tan⁡ π3=3⇒ ex=1=3 i.e. ex=4⇒x=log⁡ 4=2log2

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Value of $x$ satisfying $\int_{\log 2}^{x} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6}$ is

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log 2

3 log 2

2 log 2

answer is C.

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Detailed Solution

Substituting $e^{x} - 1 = t^{2}, e^{x} d x = 2 t d t$

$\begin{array}{l} d x = \frac{2 t d t}{t^{2} + 1}, so \\ \frac{π}{6} = \int_{\log 2}^{x} \frac{d x}{\sqrt{e^{x} - 1}} = \int_{1}^{\sqrt{e^{x} - 1}} \frac{2 t d t}{(t^{2} + 1) t} \\ = 2 \int_{1}^{\sqrt{e^{x} - 1}} \frac{d t}{t^{2} + 1} = {2 \tan^{- 1} t|}_{1}^{\sqrt{e^{x} - 1}} \\ = 2 \tan^{- 1} \sqrt{e^{x} - 1} - \frac{π}{2} \\ \Rightarrow 2 \tan^{- 1} \sqrt{e^{x} - 1} = \frac{2 π}{3} \\ \Rightarrow \sqrt{e^{x} - 1} = \tan \frac{π}{3} = \sqrt{3} \\ \Rightarrow e^{x} = 1 = 3 i.e. e^{x} = 4 \Rightarrow x = \log 4 = 2 \log 2 \end{array}$