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If-2x-y-8-and-x-y-3-2-log-10-2-log-8-36-then-x-2-3y-a-28-b-22-c-20-d-16-e-12-

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Question Number 66354 by gunawan last updated on 13/Aug/19
If 2x+y=8 and (x+y)=(3/2)log_(10) 2.log_8 36 then x^2 +3y=... a. 28 b. 22 c. 20 d. 16 e. 12
If2x+y=8and(x+y)=32log102.log836thenx2+3y=…a.28b.22c.20d.16e.12
Commented by kaivan.ahmadi last updated on 13/Aug/19
(3/2)log_(10) 2.log_8 36=(3/2)log_(10) 2.(2/3)log_2 6= log_(10) 2.log_2 6=((log_2 6)/(log_2 10))=log_(10) 6=x+y⇒ { ((2x+y=8)),((x+y=log_(10) 6)) :}⇒ { ((2x+y=8)),((−2x−2y=−2log_(10) 6)) :}⇒ y=−8+2log_(10) 6 and 2x−8+2log_(10) 6=8⇒x=8−log_(10) 6 so x^2 +3y=64−16log_(10) 6+(log_(10) 6)^2 −24+6log_(10) 6= (log_(10) 6)^2 −10log_(10) 6+40=
32log102.log836=32log102.23log26=log102.log26=log26log210=log106=x+y⇒{2x+y=8x+y=log106⇒{2x+y=8−2x−2y=−2log106⇒y=−8+2log106and2x−8+2log106=8⇒x=8−log106sox2+3y=64−16log106+(log106)2−24+6log106=(log106)2−10log106+40=

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