log2(3),log4(5),3/2怎么比较大小?
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log2(3),log4(5),3/2怎么比较大小?
log2(3),log4(5),3/2怎么比较大小?
log2(3),log4(5),3/2怎么比较大小?
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)
log2(20)-log2(5)+2log3(2)×log4(3)
log2(3),log4(5),3/2怎么比较大小?
(log2 5+log4 125)log3 2/ log根号3 5
log2 20---log2 5+log3 4 * log4 3 等于什么
log的计算 :log2(20)-log4(25)= log3(2)×log2(5)×log5(3)=?log2[log2(32)-log2(3÷4)+log2log的计算 :log2(20)-log4(25)= log3(2)×log2(5)×log5(3)=?log2【log2(32)-log2(3÷4)+log2(6)】=?lgx+lg(x+15)=2 ..X
求解log2 (3X+2)-log4 (X)=3
解方程:log2(3-x) - log4(x+5) =1
比较log2 3,log4 5,log7 6的大小
2^(2log2的3次方+log4的16次方 )
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3log4 5与2log2 3哪个比较大比较对数函数的大小
(log2(5)+log4(125))×log3(2)/log根号3(5)
(log2^5+log4^125)(log3^2)/(log(根号3)^5 要具体的过程,
(log2(5)+log4(125))(log3(2)/log√3(5))怎么解?
求值:3^(log3 (25) )+(-C(5,4)^(log2 1)+16^(log4(2sin(π/2))=
运用对数换底公式化简log2(3)*log3(4)*log4(5)*log5(2),
log2 3 · log3 4 ·log4 5 ·log5 2 前面的是底数
帮忙换底化简下列:logaC*logcA .log2 3 * log3 4 * log4 5 * log5 2