Prove that the following equations are equivalent,
$2^{2^{x-1}}=\frac{1}{2^{2^{x}}-1}$ and $2^{2^{x+1}}=\frac{1}{2^{2^{x-1}}-1}$
They both have an approximate solution of -.30157 when put in a calculator, but I'm completely how to show they are equal through changing their forms or using log properties. Any hints or suggestions?
Denote $z = 2^{2^{x-1}}$, then we have to show equivalence of the equations $$ z = \frac{1}{z^2 - 1}, \tag1 $$ $$ z^4 = \frac{1}{z - 1}, \tag2 $$ or after simple rearranging $$ z^3 - z - 1 = 0, \tag1 $$ $$ z^5 - z^4 - 1 = 0. \tag2 $$ More precisely, we want to see that these equations have the same real roots (because we assume $z = 2^{2^{x-1}}$ is real for real $x$). One may see that $$ z^5 - z^4 - 1 = (z^3 - z - 1)(z^2 - z + 1), $$
and since polynomial $z^2 - z + 1$ has no real roots, we conclude that $(1)$ has the same real roots as $(2)$.