This is the equation I am trying to solve:
$$ p=\frac{\ln(n+1)}{\ln(2n+1)} $$
I am trying to work out n in terms of p. However, this is where I am stuck:
$$ p\times\ln(2n+1)=\ln(n+1)\\ e^{p\times\ln(2n+1)}=e^{\ln(n+1)}\\ (e^{\ln(2n+1)})^p=n+1\\ (2n+1)^p=n+1\\ ??? $$
I'm not sure how to proceed from here; any suggestions would be welcome.
Update**: I got horribly confused by terminology - $p\in\mathbb{R}$ and $n\in\mathbb{R}$. Didn't realise that natural numbers are integers.
For $n>0$ we have that
$$n+1 < 2n + 1 \leq (2n+1)^p$$
with $ p > 0$. The strict inequalities means that they will never be equal. However $n=0$ is not possible because that makes the original statement undefined (it fails the "vertical line test" since $n=0$ could make $p$ be any value we want). Therefore the given expression has no solutions.