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[函数] 一道方程

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longzaifei Post time 2015-2-12 10:39 |Read mode
$\begin{cases}
2^x+2x=5 \\
2y+\log_2(y-1) =5\end{cases}$ 求 $x+y  $

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Tesla35 Post time 2015-8-25 16:16
奎宁个@kuing用软件先算一下这题

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isee Post time 2015-8-25 16:55
本帖最后由 isee 于 2015-8-25 17:14 编辑 回复 1# longzaifei


    这个如何解?若是将y-1退化成y还是不好处理

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kuing Post time 2015-8-25 17:03
回复 2# Tesla35

约等于 3.59

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Tesla35 Post time 2015-8-25 17:03
回复 4# kuing

没有精确解?

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kuing Post time 2015-8-25 17:07
回复 5# Tesla35

木有

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isee Post time 2015-9-1 16:27
楼主此题的原题是什么?

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其妙 Post time 2015-9-1 23:35
回复 7# isee
多了系数2

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kuing Post time 2015-9-2 02:40
回复 8# 其妙

既然知道原题,具体写一下,也可以复习一下代码

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isee Post time 2015-9-2 15:02
回复 9# kuing


    他写估计也是在 mathtype 里复制过来。。。。他一想兴起,可能这里帖下。。。

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kuing Post time 2015-9-4 15:39
回复 10# isee

或许就被你这么一说他干脆就不理了

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isee Post time 2015-9-4 19:13
回复  isee
多了系数2
其妙 发表于 2015-9-1 23:35



    第几个多了个2?

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 Author| longzaifei Post time 2015-9-5 19:18
不好意思,题目从artofproblemsolving复制过来的,应该是原贴有问题,时间比较久,原帖也找不到了

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tommywong Post time 2015-9-5 22:40
zhidao.baidu.com/link?url=xdoEY7m3mULfNyAr6bKYfMp_bwiaB9LHqTVlLU ... T57v_t9644L9HjB6bVqa

∵2X+2^x=5
∴x+2^(x-1)=2.5
∵2X+2log2(x-1)(2是底数)=5
∴x+log2(x-1)=2.5
其中y=2^(x-1)与y=log2(x-1)关于y=x-1对称
∴y=2.5-x与前面指数、对数曲线交点即为x1,x2.
x1+x2为y=2.5-x,y=x-1交点横坐标2倍,即为3.5

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kuing Post time 2015-9-6 03:00
擦,原来是少了系数2……

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isee Post time 2015-9-6 14:29
其妙真逗,反正结果还是对称性……

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kuing Post time 2015-9-6 14:32
回复 16# isee

这种题不能整成对称的话能做的可能性就不大了……除非是经过精心设计的恰好有解的

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