Question

Express the following in logarithmic form
$\cos h^{-1}2$

Answer 1

Let $cos{h}^{-1}2=x$

$\Rightarrow coshx=2$

$\Rightarrow \frac{e^x+e^{-x}}{2}=2$

$\Rightarrow e^{2x}-4e^x+1=0$

$\Rightarrow e^x=2\pm \sqrt{3}$

$\Rightarrow x=ln(2\pm \sqrt{3})$