Question Number 38099 by Cheyboy last updated on 21/Jun/18
$${x}^{{x}} =\mathrm{0}.\mathrm{25} \\ $$$${find}\:{x} \\ $$
Commented by MrW3 last updated on 22/Jun/18
$${no}\:{real}\:{solution},\: \\ $$$${since}\:{x}^{{x}} \:{always}\:\geqslant\frac{\mathrm{1}}{\:^{{e}} \sqrt{{e}}}=\mathrm{0}.\mathrm{6922}>\mathrm{0}.\mathrm{25} \\ $$$${see}\:{Q}\mathrm{37414} \\ $$
Commented by MrW3 last updated on 22/Jun/18
Commented by prof Abdo imad last updated on 22/Jun/18
![(e)⇔e^(xlnx) =4^(−1) =e^(−ln(4)) =e^(−2ln(2)) ⇔xln(x)+2ln(2)=0 let f(x)=xln(x)+2ln(2) with x>0 f^′ (x)=ln(x) +1 ≥0 ⇔ln(x)≥ln((1/e)) ⇔x≥(1/e) so f is increasing on [(1/e),+∞[ and decreasing on]0,(1/e)] f((1/e))=−(1/e) +2ln(2)>0 lim_(x→0^+ ) f(x)=2ln(2)>0 lim_(x→+∞) f(x)=+∞ so f(x)≠0 ∀x>0 so the equation haven t any real solution.](Q38159.png)
$$\left({e}\right)\Leftrightarrow{e}^{{xlnx}} =\mathrm{4}^{−\mathrm{1}} ={e}^{−{ln}\left(\mathrm{4}\right)} ={e}^{−\mathrm{2}{ln}\left(\mathrm{2}\right)} \:\Leftrightarrow{xln}\left({x}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${let}\:{f}\left({x}\right)={xln}\left({x}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right)\:{with}\:{x}>\mathrm{0} \\ $$$${f}^{'} \left({x}\right)={ln}\left({x}\right)\:+\mathrm{1}\:\geqslant\mathrm{0}\:\Leftrightarrow{ln}\left({x}\right)\geqslant{ln}\left(\frac{\mathrm{1}}{{e}}\right)\:\Leftrightarrow{x}\geqslant\frac{\mathrm{1}}{{e}} \\ $$$${so}\:{f}\:{is}\:{increasing}\:{on}\:\left[\frac{\mathrm{1}}{{e}},+\infty\left[\:{and}\:{decreasing}\right.\right. \\ $$$$\left.{o}\left.{n}\right]\mathrm{0},\frac{\mathrm{1}}{{e}}\right]\:\:\:\:\:\:\:{f}\left(\frac{\mathrm{1}}{{e}}\right)=−\frac{\mathrm{1}}{{e}}\:+\mathrm{2}{ln}\left(\mathrm{2}\right)>\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{f}\left({x}\right)=\mathrm{2}{ln}\left(\mathrm{2}\right)>\mathrm{0} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty\:{so}\:{f}\left({x}\right)\neq\mathrm{0}\:\forall{x}>\mathrm{0} \\ $$$${so}\:{the}\:{equation}\:{haven}\:{t}\:{any}\:{real}\:{solution}. \\ $$
Commented by Cheyboy last updated on 22/Jun/18
$${Ooh}\:{k}\:{sir}\:{thank}\:{you},{it}\:{was}\:{givin} \\ $$$${to}\:{me}\:{but}\:{i}\:{could}\:{not}\:{solve}\:{it} \\ $$
Commented by Cheyboy last updated on 22/Jun/18
$${Thank}\:{you}\:{Mr}\:{W}\mathrm{3}\:{and}\:{prof}\:{abdo} \\ $$