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Question Number 193175 by Humble last updated on 06/Jun/23
pleasesolveforxif2x2=2x
Answered by Frix last updated on 06/Jun/23
Obviouslyx=1
2x2=2xLetx=±12et⇔t=−(ln(±x)+ln22)e−2t=2±12et−2t=±ln22ettet=∓ln222t=W(∓ln222)t≈{.200534631257−2.19024886644∨−.34657359028⇒x={≈−.578620636081≈6.31972235584∨1
Answered by aba last updated on 06/Jun/23
x=1issolution2x2−2x=0⇒(2x)2−((2)x)2=0⇒(2x−(2)x)(2x+(2)x)=0⇒2x−(2)x=0∨2x+(2)x=0⇒x.(2)−x=12∨x.(2)−x=−12⇒−xln(2).e−xln(2)=−2ln(2)2∨−xln(2).e−xln(2)=2ln(2)2⇒−xln(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)≈6.31972∨x=−2ln(2)W(ln(2)22)≈−0.578621
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