Question Number 212645 by efronzo1 last updated on 20/Oct/24
$$\:\:\begin{cases}{\mathrm{x}=\mathrm{2}+\:\mathrm{log}\:_{\mathrm{2}} \mathrm{log}\:_{\mathrm{2}} \mathrm{y}}\\{\mathrm{y}=\mathrm{2}\:\mathrm{log}\:_{\mathrm{2}} \mathrm{z}\:}\\{\mathrm{z}=\mathrm{2}+\:\mathrm{log}\:_{\mathrm{2}} \:\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:}\end{cases} \\ $$