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Question Number 108240 by mathdave last updated on 15/Aug/20
Answered by john santu last updated on 15/Aug/20
◻JS△★2a+4=6⇒2a+4=2.3⇒2a+3=3...(i)32b+1=3.22⇒32b=22;3b=2then(2a+3)2b=2222ab+6b=22⇒ab+3b=1⇒ab−1=−3b;ab−2=−3b−1weobtain3ab−2=3−3b−1=1(3b)3×33ab−2=124.
Answered by 1549442205PVT last updated on 16/Aug/20
2a+4=6⇒(a+4)ln2=ln(2.3)=ln2+ln3⇒aln2=ln3−3ln2=ln38⇒a=ln38ln2(1)32b+1=12⇒(2b+1)ln3=ln(3.4)⇔2bln3+ln3=ln3+ln4⇒2bln3=ln4⇒b=ln42ln3=2ln22ln3=ln2ln3(2)From(1)(2)wegetab=ln38ln2×ln2ln3=ln38ln3=log338⇒ab−2=log338−2=log338−log332=log3(38/9)=log3(124).Therefore3ab−2=3log3(124)=124(sincealogaN=N)ConsequentlywechooseC
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