Question Number 131430 by mathlove last updated on 04/Feb/21
$$\sqrt[{\frac{\mathrm{1}}{{x}}}]{\left(\sqrt[{\frac{\mathrm{1}}{{x}}}]{\left.{x}\:^{{x}} \right)^{{x}} }\right.}=\mathrm{4}\:\:\:\:\:\:\:{find}\:\:{x}=? \\ $$
Commented by math35 last updated on 04/Feb/21
![((((x^x^2 )^(1/(1/x)) ))^(1/(1/x)) =4 note:(y)^(1/(1/n)) =y^(1/(((1/n)))) =y^n → (((x^x^2 )^x ))^(1/(1/x)) =4 (x^x^3 )^x =4 x^x^4 =4 taking ln of both sides x^4 lnx=ln4 e^(lnx^4 ) .lnx=ln4 e^(4lnx) .4lnx=4ln4 e^(4lnx) .4lnx=e^(ln4) .ln4 taking lambert W function of both sides W[e^(4lnx) .4lnx]=W[e^(ln4) .ln4] 4lnx=ln4 lnx=(1/4)ln2^2 =(2/4)ln2 lnx=(1/2)ln2 lnx=ln(√2) x=(√2)](Q131451.png)
$$\sqrt[{\frac{\mathrm{1}}{{x}}}]{\left(\sqrt[{\frac{\mathrm{1}}{{x}}}]{{x}^{{x}^{\mathrm{2}} } }\right.}=\mathrm{4} \\ $$$${note}:\sqrt[{\frac{\mathrm{1}}{{n}}}]{{y}}={y}^{\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{{n}}\right)}} ={y}^{{n}} \\ $$$$\rightarrow\:\:\sqrt[{\frac{\mathrm{1}}{{x}}}]{\left({x}^{{x}^{\mathrm{2}} } \right)^{{x}} }=\mathrm{4} \\ $$$$\:\:\:\:\:\left({x}^{{x}^{\mathrm{3}} } \right)^{{x}} =\mathrm{4} \\ $$$$\:\:\:\:\:\:{x}^{{x}^{\mathrm{4}} } =\mathrm{4} \\ $$$${taking}\:{ln}\:{of}\:{both}\:{sides} \\ $$$${x}^{\mathrm{4}} {lnx}={ln}\mathrm{4} \\ $$$${e}^{{lnx}^{\mathrm{4}} } .{lnx}={ln}\mathrm{4} \\ $$$${e}^{\mathrm{4}{lnx}} .\mathrm{4}{lnx}=\mathrm{4}{ln}\mathrm{4} \\ $$$${e}^{\mathrm{4}{lnx}} .\mathrm{4}{lnx}={e}^{{ln}\mathrm{4}} .{ln}\mathrm{4} \\ $$$${taking}\:{lambert}\:{W}\:{function}\:{of}\:{both}\:{sides} \\ $$$${W}\left[{e}^{\mathrm{4}{lnx}} .\mathrm{4}{lnx}\right]={W}\left[{e}^{{ln}\mathrm{4}} .{ln}\mathrm{4}\right] \\ $$$$\mathrm{4}{lnx}={ln}\mathrm{4} \\ $$$${lnx}=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{2}^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{4}}{ln}\mathrm{2} \\ $$$${lnx}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2} \\ $$$${lnx}={ln}\sqrt{\mathrm{2}} \\ $$$${x}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\: \\ $$
Commented by mathlove last updated on 05/Feb/21
$${what}\:{is}\:{the}\:{W}\:{function}? \\ $$
Answered by talminator2856791 last updated on 04/Feb/21
$$\:{x}\:=\:\mathrm{4} \\ $$
Commented by mathlove last updated on 04/Feb/21
$${pleas}\:{solve} \\ $$
Answered by mr W last updated on 05/Feb/21
$${i}\:{interpret}\:{the}\:{question}\:{as}: \\ $$$$\sqrt[{\frac{\mathrm{1}}{{x}}}]{\left(\sqrt[{\frac{\mathrm{1}}{{x}}}]{{x}^{{x}} }\right)^{{x}} }=\mathrm{4} \\ $$$$\Leftrightarrow\sqrt[{\frac{\mathrm{1}}{{x}}}]{\left(\left({x}^{{x}} \right)^{{x}} \right)^{{x}} }=\mathrm{4} \\ $$$$\Leftrightarrow\left(\left(\left({x}^{{x}} \right)^{{x}} \right)^{{x}} \right)^{{x}} =\mathrm{4} \\ $$$$\Leftrightarrow\left(\left({x}^{{x}^{\mathrm{2}} } \right)^{{x}} \right)^{{x}} =\mathrm{4} \\ $$$$\Leftrightarrow\left({x}^{{x}^{\mathrm{3}} } \right)^{{x}} =\mathrm{4} \\ $$$$\Leftrightarrow{x}^{{x}^{\mathrm{4}} } =\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{4}} =\mathrm{4} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{4}}]{\mathrm{4}}=\sqrt{\mathrm{2}}>\mathrm{0} \\ $$
Commented by mathlove last updated on 05/Feb/21
$${thanks}\:{a}\:{lot}\:{sir} \\ $$
Commented by mr W last updated on 05/Feb/21
$${x}^{{x}^{{x}^{\iddots^{{x}^{{a}} } } } } =\:{a} \\ $$$${has}\:{the}\:{same}\:{solution}\:{as}\:{x}^{{a}} ={a}, \\ $$$${i}.{e}.\:{x}=\sqrt[{{a}}]{{a}} \\ $$
Commented by Tawa11 last updated on 23/Jul/21
$$\mathrm{great} \\ $$