Question

If $p\in R$ and $q={\log }_x(p+\sqrt{p^2+1})$, then find the value of p in terms of x and q.

Answer 1

Given,

$q={\log }_x(p+\sqrt{p^2+1})$

$x^q=p+\sqrt{p^2+1}$

$x^{-q}=\frac{1}{x^q}=\frac{1}{p+\sqrt{p^2+1}}$

$=\frac{p-\sqrt{p^2+1}}{(p+\sqrt{p^2+1})(p-\sqrt{p^2+1})}$

$=\frac{p-\sqrt{p^2+1}}{p^2-(p^2+1)}=-p+\sqrt{p^2+1}$

Therefore,

$x^q-x^{-q}=p+\sqrt{p^2+1}-\sqrt{p^2+1}+p$

$x^q-x^{-q}=2p$

$\frac{x^q-x^{-q}}{2}=p$