\(\sqrt{x}\) = y --- we can now square bothside
\(x = y^2\) --- square two more times on bothsides
\(x^4 = y^8\)
\(\frac{1}{(x^{−2})^{−2}}\) =\(\frac{1}{x^{4}}\) =\(\frac{1}{y^{8}}\) =\(y^{-8}\)
...
\(x = y^2\) --- square two more times on bothsides
\(x^4 = y^8\)
\(\frac{1}{(x^{−2})^{−2}}\) =\(\frac{1}{x^{4}}\) =\(\frac{1}{y^{8}}\) =\(y^{-8}\)
[Reveal] Spoiler:
A
...
![FLO – Therapy at the Club – Pre-Single [iTunes Plus M4A]](http://is1-ssl.mzstatic.com/image/thumb/Music221/v4/a4/11/88/a41188ed-0f2e-e8d5-1163-d59e373cc58c/26UMGIM30046.rgb.jpg/208x208bb.webp)
.jpg)
