fyqt.net
当前位置:首页 >> log2(3)*log3(4)=? >>

log2(3)*log3(4)=?

log2(3)*log3(4)=log2(3)*log2(4)/log2(3)=log2(4)=2

Log3(4)=lg4/lg3=2lg2/lg3=2x0.3010/0.4771=1.2617899811≈1.262

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

(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

都换成以2为底的对数进行比较 log2(3)=log2(根号9) log4(5)=log2(根号5) 3/2=log2(2^(3/2))=log2(根号8) 因为y=log2(x)在(0,+无穷)上为增函数 所以log2(3)>3/2>log4(5)

log4(3)+log8(3) =log2(3)/log2(4)+log2(3)/log2(8) =0.5log2(3)+1/3*log2(3) =5/6*log2(3) 在高中阶段运算到这一步化为单个对数就可以了,主要运用的是换底公式。

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)...

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

(log2^3+log4^9)(log3^4+log9^2)= (3*log2+18*log2)(4*log3+4*log3)= 21*log2*8*log3=168log5

log4(6)=1/2log2(6)=log2(√6)log4(6)

网站首页 | 网站地图
All rights reserved Powered by www.fyqt.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com