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If $\cosh^{-1} x=2 \log_{e} (\sqrt{2}+1),$ then x= 

Answer:

Explanation:

We have,   $\cosh^{-1} x=2 \log_{e} (\sqrt{2}+1),$

$\Rightarrow$   $\log_{e}(x+\sqrt{x^{2}-1})= \log_{e}(\sqrt{2}+1)^{2}$

$\Rightarrow$   $x+\sqrt{x^{2}-1}= (\sqrt{2}+1)^{2}$

$\Rightarrow$     $x+\sqrt{x^{2}-1}= 3+2\sqrt{2}=3+\sqrt{8}$

 On comparing rational and irrational part, we get

   x=3