求一log题20=X/log(1100/33),请问X是多少?写错了,应该是20=X/2log(1100/33)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/26 22:17:57
求一log题
20=X/log(1100/33),请问X是多少?
写错了,应该是20=X/2log(1100/33)
20=X/log(1100/33),请问X是多少?
写错了,应该是20=X/2log(1100/33)
20=X/2log(1100/33)
x=40log(1100/33)
=40(lg1100-lg33)
=40[lg11+lg100-1g33]
=40[lg11/33+2]
=40[lg1/3+2]
=40[-lg3+2]
=40[2-lg3]
=40[2-0.4771]
=60.916
x=40log(1100/33)
=40(lg1100-lg33)
=40[lg11+lg100-1g33]
=40[lg11/33+2]
=40[lg1/3+2]
=40[-lg3+2]
=40[2-lg3]
=40[2-0.4771]
=60.916
[( log(x^2) + log(x^4) )/ log(100 x)] = 4 求X
log₂log₃log₄ X=log₃log₄logS
已知log(2)[log(3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x
2 log(3x)= log 4 +log(6-x)怎么求...急
(高一)若x满足2(log(1/2)x)^2-14log(4)x+3≤0,求f(x)=[log(2)(x/2)]*{lo
log(0.5x)(2)-log(0.5x^3)(x^2)=log(0.5x^3)(4)
solve:log(x)+log(2x)=2
log(X+5)+log(X+2)=1
x=log a 2
一道高一数学题:解方程:log(0.5x)(2)-log(0.5x^3)(X^2)=log(0.5x^3)(4) 求大神
2log(b)x=log(a)x+log(c)x
数学log题x是多少.