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Q.
If $\cos h^{-1} x = 2 \log_e ( \sqrt{2} + 1 )$, then x =
TS EAMCET 2017
Solution:
We have, $\cosh ^{-1} x=2 \log _{\theta}(\sqrt{2}+1)$
$\Rightarrow \log _{e}\left(x+\sqrt{x^{2}-1}\right)=\log _{e}(\sqrt{2}+1)^{2}$
$\Rightarrow x+\sqrt{x^{2}-1}=(\sqrt{2}+1)^{2}$
$\Rightarrow x+\sqrt{x^{2}-1}=3+2 \sqrt{2}=3+\sqrt{8}$
On comparing rational and irrational part, we get
$x=3$