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x-x-0-25-find-x-




Question Number 38099 by Cheyboy last updated on 21/Jun/18
x^x =0.25  find x
xx=0.25findx
Commented by MrW3 last updated on 22/Jun/18
no real solution,   since x^x  always ≥(1/(^e (√e)))=0.6922>0.25  see Q37414
norealsolution,sincexxalways1ee=0.6922>0.25seeQ37414
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.
(e)exlnx=41=eln(4)=e2ln(2)xln(x)+2ln(2)=0letf(x)=xln(x)+2ln(2)withx>0f(x)=ln(x)+10ln(x)ln(1e)x1esofisincreasingon[1e,+[anddecreasingon]0,1e]f(1e)=1e+2ln(2)>0limx0+f(x)=2ln(2)>0limx+f(x)=+sof(x)0x>0sotheequationhaventanyrealsolution.
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
Oohksirthankyou,itwasgivintomebuticouldnotsolveit
Commented by Cheyboy last updated on 22/Jun/18
Thank you Mr W3 and prof abdo
ThankyouMrW3andprofabdo

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