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Question Number 106076 by Skabetix last updated on 02/Aug/20

solve this pls  x^x^(1/2)  =(1/2)

$${solve}\:{this}\:{pls} \\ $$$${x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

Another method  x^x^(1/2)  =(1/2)    (look that (1/2)=x^x^(1/2)    x^x^x^x^(1/2)    =(1/2)  like this manner  x^x^x^x^x^x^x      =(1/2)  so   x^(1/2) =(1/2)          x=(1/4)    It is general for   X^X^N  =N  X=(N)^(1/N)

$${Another}\:{method} \\ $$$${x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } =\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\left({look}\:{that}\:\frac{\mathrm{1}}{\mathrm{2}}={x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } \right. \\ $$$${x}^{{x}^{{x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } } } =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${like}\:{this}\:{manner} \\ $$$${x}^{{x}^{{x}^{{x}^{{x}^{{x}^{{x}} } } } } } =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:\:\:{x}^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${It}\:{is}\:{general}\:{for}\: \\ $$$${X}^{\mathrm{X}^{\mathrm{N}} } ={N} \\ $$$${X}=\sqrt[{\mathrm{N}}]{\mathrm{N}} \\ $$

Commented by mr W last updated on 02/Aug/20

there are two roots:  x=e^(2W(−((ln 2)/2))) = { ((1/4)),((1/(16))) :}

$${there}\:{are}\:{two}\:{roots}: \\ $$$${x}={e}^{\mathrm{2}\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)} =\begin{cases}{\frac{\mathrm{1}}{\mathrm{4}}}\\{\frac{\mathrm{1}}{\mathrm{16}}}\end{cases} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

x^(1/2) logx=log((1/2))  logx.e^((1/2)log(x)) =log((1/2))  (1/2)log(x)e^((1/2)log(x)) =(1/2)log((1/2))  (1/2)log(x)=W_0 ((1/2)log((1/2)))  x=e^(2W_0 (−log((√2))) =(1/4),(1/(16))

$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} {logx}={log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${logx}.{e}^{\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right)} ={log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right)} =\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right)={W}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${x}={e}^{\mathrm{2}{W}_{\mathrm{0}} \left(−{log}\left(\sqrt{\mathrm{2}}\right)\right.} =\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{16}} \\ $$

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