Question

Show that $\log [\sqrt{x^2+1}+x]+\log [\sqrt{x^2+1}-x]=0$

Answer 1

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

$=\log \left(\right(\sqrt{x^2+1}+x\left)\right(\sqrt{x^2+1}-x\left)\right)$

$=\log \left(\right(\sqrt{x^2+1}{)}^2-\left(x{)}^2\right)$

$=\log (x^2+1-x^2)=\log 1=0$

hence proved