若log(1+k)^(1-k)
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若log(1+k)^(1-k)
若log(1+k)^(1-k)<1则实数k的取值范围
若log(1+k)^(1-k)
首先明确定义域:1+k>0且1-k>0解得-1
底数是多少呀?
若log(1+k)^(1-k)
若log(1+k) (1-k)
请问为什么1/log(a)k=log(k)a?
求大神指点,lingo出现错误代码11,这是为什么?model:!目标函数;min=(0.005*((@log(1+k))^2+(@log(1+2*k))^2+(@log(1+3*k))^2+(@log(1+4*k))^2+(@log(1+5*k))^2))!/(@log(1+k)+@log(1+2*k)+@log(1+3*k)+@log(1+4*k)+@log(1+5*k));!约束条件;(@log
那么为什么log(k+1)>log(k+2)呢 能写清楚些吗
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