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(log2 6)×(log3 6)%(log2 3+log3 2)=

log2(3)-log4(6) = log2(3)-log2^2((√6)^2) = log2(3)-log2(√6) = log2(3/√6) = log2(√3/√2) = log2(√3)-log2(√2) = 1/2log2(3) - 1/2 如果题目是【log2(3)-log4(36)】 log2(3)-log4(36) = log2(3)-log2^2((6)^2) = log2(3)-log2(6) = log2(3/6)...

设所有等式的值为t 所以a=2^(t-2) b=3^(t-3) a+b=6^t 所以1/a+1/b=(a+b)/ab=6^t/[2^(t-2)*3^(t-3)]=6^t/[(2^t)/4*(3^t)/27]=108

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

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俊狼猎英团队为您解答 已知lg2=a,lg3=b, log3 4=lg4/lg3=2lg2/lg3=2a/b, log2 12=log2 (2^2×3)=2+log2 3=2+lg3/lg2=2+b/a。 前面部分意思不懂。

log₃8-2log₃6 =log₃2³-2(log₃3+log₃2) =3log₃2-2(1+log₃2) =log₃2 -2 =a-2

你参考看看!

(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!

解 log2(9)×log3(8) =log2(3²)×log3(2³) =2log2(3)×3log3(2) =6×log2(3)×log3(2) =6×1 =6

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