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Question Number 160976 by alf123 last updated on 10/Dec/21
Answered by TheSupreme last updated on 10/Dec/21
xy=s2sy+3y=212s2=3→s=2ln(3)ln(2)32y+3y=21t2+3t=21t=−3±9+842=−3+932y=ln(3)[93−32]x=ln2(3)ln(2)[93−3]1x+1y=193−3[ln(2)ln2(3)+1ln(3)]
Answered by FelipeLz last updated on 10/Dec/21
{2x−3y=212xy=32xy=3⇒(2xy)12=3⇒2xy=32⇒2x=32y2x−3y=21⇒32y−3y−21=03y=u⇒u2−u−21=0u=1±1+4⋅212=1±8523y=1±852⇒y=log3(1±852)=ln(1±85)−ln(2)ln(3)2x=32y⇒xln(2)=2yln(3)⇒xln(2)=2ln(3)[ln(1±85)−ln(2)ln(3)]⇒x=2[ln(1±85)ln(2)−1]=2log2(1±852){x=2log2(1±852)y=log3(1±852)1±852=a1x+1y=12log2(a)+1log3(a)=12loga(2)+loga(3)=loga(2)+loga(3)=loga(32)1x+1y=log(1±852)(32)
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