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(log2 3+log4 27)(log3 4+log9 8)

35/4

过程如图,

loga(b)*logb(c)=loga(c); loga(b)*logb(c)*logc(d)=loga(d). log2(3)*log3(4)*log4(5)*log5(6)*log6(7)*log7(8)=log2(8)=3,所以3=loga(b),所以b=a^3

(log3 2+log9 2)*(log4 3+log8 3) =(lg2/lg3+lg2/lg9)*(lg3/lg4+lg3/lg8) =(lg2/lg3+lg2/2lg3)*(lg3/2lg2+lg3/3lg2 =3/2*lg2/lg3*5/6*lg3/lg2 =3/2*5/6=5/4

我会,就是把4变为二的平方,所以就是(6log2+3log8) (2log3+4log3)再分别相乘,得出结果!

(log4(3)+log8(3))(log3(2)+log9(2)) =(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9) =5lg3/6lg2X3lg2/2lg3 =5/6X3/2 =5/4

(log3² log9²)(log4³ log8³) =(lg2/lg3*lg2/[2lg3])*(lg3/[2lg2]*lg3/[3lg2]) =(1/2)*(1/[2]*1/[3]) =1/12

(log4(3)+iog8(3))(iog3(2)+iog9(2))-iog1/2(32^(1/4)) =[1/2log2 3+1/3log3 2][log3 2+1/2log3 2]-log1/2 [2^(5/4)] =5/6log2 3*3/2log3 2-5/4 =5/4-5/4=0

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