Question Number 221754 by fantastic last updated on 09/Jun/25
$$\left.\sqrt[{\sqrt[{\sqrt[{\:\mathrm{3}^{\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } } }]{\mathrm{27}}}]{\mathrm{64}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$
Answered by wewji12 last updated on 09/Jun/25
$$…..\mathrm{what}\:\mathrm{a}\:\mathrm{horrible}\:\mathrm{notation} \\ $$$${A}=\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } =\mathrm{1}\:\therefore\mathrm{3}^{{A}} =\mathrm{3} \\ $$$$\:^{\sqrt{\mathrm{27}}} \:^{\sqrt{\mathrm{64}}} \sqrt{\mathrm{81}}=\:^{\:^{\mathrm{3}\sqrt{\mathrm{3}}} \mathrm{8}} \mathrm{9}=\mathrm{9}^{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{3}\sqrt{\mathrm{3}}} } =\mathrm{9}^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} } \\ $$$$\sqrt{\mathrm{4}}\:\mathrm{blank}\:^{\mathrm{24}−\Sigma{n}} \sqrt{\mathrm{81}}….??\: \\ $$