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Question Number 193175 by Humble last updated on 06/Jun/23

please solve for x if  2x^2 =2^x

pleasesolveforxif2x2=2x

Answered by Frix last updated on 06/Jun/23

Obviously x=1

Obviouslyx=1

Answered by Frix last updated on 06/Jun/23

2x^2 =2^x   Let x=±(1/( (√2)e^t )) ⇔ t=−(ln (±x)+((ln 2)/2))  e^(−2t) =2^(±(1/( (√2)e^t )))   −2t=±((ln 2)/( (√2)e^t ))  te^t =∓((ln 2)/(2(√2)))  t=W(∓((ln 2)/(2(√2))))  t≈ { ((.200534631257)),((−2.19024886644∨−.34657359028)) :}  ⇒  x= { ((≈−.578620636081)),((≈6.31972235584∨1)) :}

2x2=2xLetx=±12ett=(ln(±x)+ln22)e2t=2±12et2t=±ln22ettet=ln222t=W(ln222)t{.2005346312572.19024886644.34657359028x={.5786206360816.319722355841

Answered by aba last updated on 06/Jun/23

x=1 is solution  2x^2 −2^x =0 ⇒ ((√2)x)^2 −(((√2))^x )^2 =0 ⇒ ((√2)x−((√2))^x )((√2)x+((√2))^x )=0  ⇒(√2)x−((√2))^x =0 ∨ (√2)x+((√2))^x =0  ⇒x.((√2))^(−x) =(1/( (√2))) ∨ x.((√2))^(−x) =−(1/( (√2)))  ⇒−xln((√2)).e^(−xln((√2))) =−(((√2)ln((√2)))/2) ∨ −xln((√2)).e^(−xln((√2))) =(((√2)ln((√2)))/2)  ⇒−xln((√2))=W(−(((√2)ln((√2)))/2))∨−xln((√2))=W((((√2)ln((√2)))/2))  ⇒x=−(2/(ln(2)))W(−(((√2)ln(2))/4))∨x=−(2/(ln(2)))W((((√2)ln(2))/4))  ⇒x=−(2/(ln(2)))W(−((ln(2))/(2(√2))))≈6.31972 ∨ x=−(2/(ln(2)))W(((ln(2))/(2(√2))))≈−0.578621

x=1issolution2x22x=0(2x)2((2)x)2=0(2x(2)x)(2x+(2)x)=02x(2)x=02x+(2)x=0x.(2)x=12x.(2)x=12xln(2).exln(2)=2ln(2)2xln(2).exln(2)=2ln(2)2xln(2)=W(2ln(2)2)xln(2)=W(2ln(2)2)x=2ln(2)W(2ln(2)4)x=2ln(2)W(2ln(2)4)x=2ln(2)W(ln(2)22)6.31972x=2ln(2)W(ln(2)22)0.578621

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