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Question Number 110856 by mathdave last updated on 31/Aug/20
find x in x^2 =5^x^2 −19
findxinx2=5x2−19
Answered by mr W last updated on 31/Aug/20
let t=x^2 >0 t=5^t −19 5^t =t+19 5^(t+19) =(t+19)×5^(19) e^((t+19)ln 5) =(t+19)×5^(19) 1=e^(−(t+19)ln 5) (t+19)×5^(19) −(t+19)ln 5e^(−(t+19)ln 5) =−((ln 5)/5^(19) ) ⇒−(t+19)ln 5=W(−((ln 5)/5^(19) )) ⇒t=−19−(1/(ln 5))W(−((ln 5)/5^(19) )) ⇒x=±(√(−19−(1/(ln 5))W(−((ln 5)/5^(19) ))))≈±1.3742
lett=x2>0t=5t−195t=t+195t+19=(t+19)×519e(t+19)ln5=(t+19)×5191=e−(t+19)ln5(t+19)×519−(t+19)ln5e−(t+19)ln5=−ln5519⇒−(t+19)ln5=W(−ln5519)⇒t=−19−1ln5W(−ln5519)⇒x=±−19−1ln5W(−ln5519)≈±1.3742
Commented by mathdave last updated on 31/Aug/20
perfect solution
perfectsolution
Answered by Dwaipayan Shikari last updated on 31/Aug/20
x^2 =5^x^2 −19 5^a =a+19 5^(a+19) =(a+19)5^(19) 5^(−(a+19)) =(1/((a+19)))5^(−19) −(a+19)e^(log5^(−(a+19)) ) =−5^(−19) −(a+19)log5e^(−(a+19)log5) =−log(5)5^(−19) −(a+19)log5=W_0 (−log(5)5^(−19) ) a=−19−((W_0 (−log(5).5^(−19) ))/(log5))=I x=(√I) =±1.3742
x2=5x2−195a=a+195a+19=(a+19)5195−(a+19)=1(a+19)5−19−(a+19)elog5−(a+19)=−5−19−(a+19)log5e−(a+19)log5=−log(5)5−19−(a+19)log5=W0(−log(5)5−19)a=−19−W0(−log(5).5−19)log5=Ix=I=±1.3742

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