A) log 2
B) log 3
C) log 4
D) None of these
Correct Answer: C
Solution :
Let \[l=\int\limits_{\log 2}^{x}{\frac{1}{\sqrt{{{e}^{x}}-1}}}dx\] Put \[{{e}^{x}}-1={{t}^{2}}\] \[\Rightarrow \] \[{{e}^{x}}dx=2t\,dt\] \[\therefore \] \[l=2\int\limits_{1}^{\sqrt{{{e}^{x}}-1}}{\frac{1}{1+{{t}^{2}}}dt}\] \[=2\left[ {{\tan }^{-1}}\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}1 \right]\] \[=2{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}-\frac{\pi }{4}\] But \[\int\limits_{\log _{e}^{2}}^{x}{\frac{1}{\sqrt{{{e}^{x}}-1}}}dx=\frac{\pi }{6}\] \[=2{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}\] \[=\frac{2\pi }{3}\] \[\Rightarrow \] \[\sqrt{{{e}^{x}}-1}=\tan \frac{\pi }{3}\] \[=\sqrt{3}\] \[={{e}^{x}}-1\] \[=3\] \[\Rightarrow \] \[{{e}^{x}}=4\] \[\therefore \] \[x=\log \,4\]
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