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Question Number 10248 by j.masanja06@gmail.com last updated on 31/Jan/17
solve the value of x logx^2 =(x/(25))
solvethevalueofxlogx2=x25
Answered by mrW1 last updated on 01/Feb/17
I. if x>0: log x^2 =(x/(25)) ⇒2log x=(x/(25)) log x=(x/(50)) ((ln x)/(ln 10))=(x/(50)) ln x=((ln 10)/(50))x=−ax with a=−((ln 10)/(50)) ln x=−ax x=e^(−ax) xe^(ax) =1 (ax)e^(ax) =a ⇒ax=W(a) Lambert W function ⇒x=((W(a))/a)=−((W(−((ln 10)/(50))))/((ln 10)/(50)))=−((W(−0.046052))/(0.046052)) { ((=((−0.048332)/(−0.046052))=1.049516)),((=((−4.605162)/(−0.046052))=100)) :} II. if x<0: log x^2 =(x/(25)) let t=−x, t>0 log (−t)^2 =−(t/(25)) log (t)^2 =−(t/(25)) 2log t=−(t/(25)) see above t=((W(((ln 10)/(50))))/((ln 10)/(50))) x=−t=−((W(((ln 10)/(50))))/((ln 10)/(50)))=−((W(0.046052))/(0.046052)) =−((0.044067)/(0.046052))=−0.956903 ⇒ x=−((W(±((ln 10)/(50))))/((ln 10)/(50)))≈ { ((−0.956903)),((1.049516)),((100)) :}
I.ifx>0:logx2=x25⇒2logx=x25logx=x50lnxln10=x50lnx=ln1050x=−axwitha=−ln1050lnx=−axx=e−axxeax=1(ax)eax=a⇒ax=W(a)LambertWfunction⇒x=W(a)a=−W(−ln1050)ln1050=−W(−0.046052)0.046052{=−0.048332−0.046052=1.049516=−4.605162−0.046052=100II.ifx<0:logx2=x25lett=−x,t>0log(−t)2=−t25log(t)2=−t252logt=−t25seeabovet=W(ln1050)ln1050x=−t=−W(ln1050)ln1050=−W(0.046052)0.046052=−0.0440670.046052=−0.956903⇒x=−W(±ln1050)ln1050≈{−0.9569031.049516100
Answered by arge last updated on 04/Feb/17
2log x=(x/(25)) log x=y x=50y
2logx=x25logx=yx=50y

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