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Question Number 160976 by alf123 last updated on 10/Dec/21

Answered by TheSupreme last updated on 10/Dec/21

(x/y)=s  2^(sy) +3^y =21  2^(s/2) =3 → s = 2 ((ln(3))/(ln(2)))  3^(2y) +3^y =21  t^2 +3t=21  t=((−3±(√(9+84)))/2)=((−3+(√(93)))/2)  y=ln(3)[(((√(93))−3)/2)]  x=((ln^2 (3))/(ln(2)))[(√(93))−3]  (1/x)+(1/y)=(1/( (√(93))−3))[((ln(2))/(ln^2 (3)))+(1/(ln(3)))]

xy=s2sy+3y=212s2=3s=2ln(3)ln(2)32y+3y=21t2+3t=21t=3±9+842=3+932y=ln(3)[9332]x=ln2(3)ln(2)[933]1x+1y=1933[ln(2)ln2(3)+1ln(3)]

Answered by FelipeLz last updated on 10/Dec/21

 { ((2^x −3^y  = 21)),(((√2^(x/y) ) = 3)) :}  (√2^(x/y) ) = 3 ⇒ (2^(x/y) )^(1/2)  = 3 ⇒ 2^(x/y)  = 3^2  ⇒ 2^x  = 3^(2y)    2^x −3^y  = 21 ⇒ 3^(2y) −3^y −21 = 0  3^y  = u ⇒ u^2 −u−21 = 0  u = ((1±(√(1+4∙21)))/2) = ((1±(√(85)))/2)  3^y  = ((1±(√(85)))/2) ⇒ y = log_3 (((1±(√(85)))/2)) = ((ln(1±(√(85)))−ln(2))/(ln(3)))  2^x  = 3^(2y)  ⇒ xln(2) = 2yln(3) ⇒ xln(2) = 2ln(3)[((ln(1±(√(85)))−ln(2))/(ln(3)))] ⇒ x = 2[((ln(1±(√(85))))/(ln(2)))−1] = 2log_2 (((1±(√(85)))/2))   { ((x = 2log_2 (((1±(√(85)))/2)))),((y = log_3 (((1±(√(85)))/2)))) :}  ((1±(√(85)))/2) = a  (1/x)+(1/y) = (1/(2log_2 (a))) + (1/(log_3 (a))) = (1/2)log_a (2)+log_a (3) = log_a ((√2))+log_a (3) = log_a (3(√2))  (1/x)+(1/y) = log_((((1±(√(85)))/2))) (3(√2))

{2x3y=212xy=32xy=3(2xy)12=32xy=322x=32y2x3y=2132y3y21=03y=uu2u21=0u=1±1+4212=1±8523y=1±852y=log3(1±852)=ln(1±85)ln(2)ln(3)2x=32yxln(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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