Question Number 72339 by naka3546 last updated on 27/Oct/19
$$\sqrt{{x}\:+\:\sqrt{\mathrm{4}{x}\:+\:\sqrt{\mathrm{16}{x}\:+\:…\:+\:\sqrt{\mathrm{4}^{\mathrm{2019}} {x}\:+\:\mathrm{3}}}}}\:\:=\:\:\sqrt{{x}}\:+\:\mathrm{1} \\ $$
Commented by naka3546 last updated on 27/Oct/19
$${x}\:\:=\:\:? \\ $$