Prove that the following equations are equivalent, for $x$ belonging to all real numbers.
$2^{2x-1}=\frac{1}{2^{2x}-1}$ and $2^{2x+1}=\frac{1}{2^{2x-1}-1}$
My work: With the left equation I see $2^{2x-1}=\frac{1}{4^x-1}=\frac{1}{(2^x-1)(2^x+1)}$ and right equation is $2^{2x+1}=\frac{1}{4^x-2}$. Maybe logs are helpful but can't see the log properties that would be useful.
The equations $$2^{2x-1}=\frac{1}{2^{2x}-1}$$ and $$2^{2x+1}=\frac{1}{2^{2x-1}-1}$$ are not equivalent.
Note that $x=1/2$ satisfies the first one but it does not satisfy the second one.
Also $x=\log_4 (1/2 +\sqrt2)$ satisfies the second one but it does not satisfy the first one.