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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△)/★)  2^(a+4)  = 6 ⇒2^(a+4)  = 2.3  ⇒2^(a+3)  = 3 ...(i)  3^(2b+1)  = 3.2^2 ⇒3^(2b)  = 2^2 ; 3^b =2  then (2^(a+3) )^(2b)  = 2^2   2^(2ab+6b)  = 2^2 ⇒ab+3b = 1   ⇒ab−1 = −3b ; ab−2=−3b−1  we obtain 3^(ab−2)  = 3^(−3b−1) =(1/((3^b )^3 ×3))  3^(ab−2)  = (1/(24)) .

JS2a+4=62a+4=2.32a+3=3...(i)32b+1=3.2232b=22;3b=2then(2a+3)2b=2222ab+6b=22ab+3b=1ab1=3b;ab2=3b1weobtain3ab2=33b1=1(3b)3×33ab2=124.

Answered by 1549442205PVT last updated on 16/Aug/20

2^(a+4) =6⇒(a+4)ln2=ln(2.3)=ln2+ln3  ⇒aln2=ln3−3ln2=ln(3/8)⇒a=((ln(3/8))/(ln2))(1)  3^(2b+1) =12⇒(2b+1)ln3=ln(3.4)⇔  2bln3+ln3=ln3+ln4⇒2bln3=ln4  ⇒b=((ln4)/(2ln3))=((2ln2)/(2ln3))=((ln2)/(ln3))(2)  From(1)(2)we get ab=((ln(3/8))/(ln2))×((ln2)/(ln3))  =((ln(3/8))/(ln3))=log_3 (3/8)  ⇒ab−2=log_3 (3/8)−2=log_3 (3/8)−log_3 3^2   =log_3 ((3/8)/9)=log_3 ((1/(24))).Therefore  3^(ab−2) =3^(log_3 ((1/(24)))) =(1/(24))(since a^(log_a N) =N)  Consequently we choose C

2a+4=6(a+4)ln2=ln(2.3)=ln2+ln3aln2=ln33ln2=ln38a=ln38ln2(1)32b+1=12(2b+1)ln3=ln(3.4)2bln3+ln3=ln3+ln42bln3=ln4b=ln42ln3=2ln22ln3=ln2ln3(2)From(1)(2)wegetab=ln38ln2×ln2ln3=ln38ln3=log338ab2=log3382=log338log332=log3(38/9)=log3(124).Therefore3ab2=3log3(124)=124(sincealogaN=N)ConsequentlywechooseC

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