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

亲,若满意该答案请采纳吧!!谢谢!!

(log2 3+log4 9)(log3 4+log9 2) =[log(2)3+4log(2)3][2log(3)2+2log(3)2] =5log(2)3x4log(3)2 =20

(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

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

解: 原式=log4(√2)-log4(2)+log9(√3)-log9(3) =1/4-1/2+1/4-1/2 =不解释了 望采纳

原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)-log2^4-log2^(√32)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3)-2-5/2=(1/2+1/3)(lg3/lg2)(1+1/2)(lg2/lg3)-2-5/2=(5/6)×(3/2)-2-5/2=5/4-2-5/2=-13/4

(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!

是×还是比大小 ×的话:原式=(lg25/lg2)×(lg4/lg3)×(lg9/lg5) =(2lg5/lg2)×(2lg2/lg3)×(2lg3/lg5) =8 比大小的话:log2 16

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