Question Number 202984 by hardmath last updated on 06/Jan/24
$$\mathrm{If}\:\:\:\mathrm{x}^{\boldsymbol{\mathrm{x}}^{\mathrm{3}} } \:=\:\mathrm{3} \\ $$$$\mathrm{Find}\:\:\:\mathrm{x}\:=\:? \\ $$
Commented by mr W last updated on 07/Jan/24
$${generally} \\ $$$${if}\:{x}^{{x}^{…^{{x}^{{a}} } } } ={a}\:>\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{{a}}]{{a}} \\ $$$${with}\:{any}\:{number}\:{of}\:\:{levels}\:{of}\:{the} \\ $$$${power}\:{tower}. \\ $$
Commented by hardmath last updated on 08/Jan/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Frix last updated on 06/Jan/24
$$\mathrm{Simply}\:\mathrm{let}\:{x}=\mathrm{e}^{\frac{{t}}{\mathrm{3}}} \:\Rightarrow \\ $$$$\mathrm{e}^{\frac{\mathrm{e}^{{t}} {t}}{\mathrm{3}}} =\mathrm{3} \\ $$$$\mathrm{e}^{{t}} {t}=\mathrm{3ln}\:\mathrm{3} \\ $$$${t}=\mathrm{ln}\:\mathrm{3} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$
Commented by Frix last updated on 06/Jan/24
$${x}^{{x}^{{n}} } ={n}\:\Rightarrow\:{x}=\sqrt[{{n}}]{{n}} \\ $$
Answered by EricAsaah last updated on 06/Jan/24
$$\left({x}^{{x}^{\mathrm{3}} } \right)^{\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}{x}^{\mathrm{3}} } =\mathrm{3}^{\mathrm{3}} \\ $$$$\left({x}^{\left.\mathrm{3}\right)^{{x}^{\mathrm{3}} } } =\mathrm{3}^{\mathrm{3}} \:\:\:\:\:\:\:\:{Let}\:{x}^{\mathrm{3}\:} =\:{t}\right. \\ $$$${t}^{{t}} =\:\mathrm{3}^{\mathrm{3}} \\ $$$$\rightarrow{t}=\mathrm{3} \\ $$$$\therefore\:{x}^{\mathrm{3}} =\mathrm{3} \\ $$$${x}=^{\mathrm{3}} \sqrt{\mathrm{3}} \\ $$$$ \\ $$
Commented by hardmath last updated on 08/Jan/24
$$\mathrm{thank}\:\mathrm{hou}\:\mathrm{dear}\:\mathrm{professors} \\ $$