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来自人教群的 $\ln^2x-[\ln x]-2=0$ 有几个解

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kuing Posted at 2013-12-8 15:47:14 |Read mode
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设 $x=e^t$,则方程化为 $t^2-2=[t]$,因为 $t\geqslant [t]>t-1$,所以 $t\geqslant t^2-2>t-1$,解得
\[
-1\leqslant t<\frac{1-\sqrt5}2~\text{或}~\frac{1+\sqrt5}2<t\leqslant 2,\]
又由 $t^2-2=[t]$ 知 $t^2$ 必为自然数,于是可设 $\abs t=\sqrt n$, $n\in\mbb N$,在上述解出的范围中找出所有符合 $\abs t=\sqrt n$ 的数为 $t=-1$, $t=\sqrt3$, $t=2$,经检验全部符合等式,所以原方程共三个解。

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其妙 Posted at 2013-12-8 22:57:44
内江市高2014届三模考试题
blog.sina.com.cn/s/blog_59a5413d0101ic6k.html

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 Author| kuing Posted at 2013-12-8 23:11:29
回复 2# 其妙

出处党牛笔……
PS、还改了数据,不过明显没区别……

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其妙 Posted at 2013-12-9 18:28:42
回复 3# kuing
还是姊妹题呢,

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isee Posted at 2013-12-10 00:31:40
第一眼,直接换元。
果然思路一致。

不过,特别奇怪,$x=e^t$,好像特殊值一般,再细一看,晕,换了个写法而已

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isee Posted at 2013-12-10 00:34:12
内江市高2014届三模考试题
其妙 发表于 2013-12-8 22:57
怎么没看见?第几题?


不过,四川的复习速度真快,全部结束了,还是复习顺序不同?

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 Author| kuing Posted at 2013-12-10 00:57:07
回复 5# isee

嗯,ln 完全是吓人,一换元就灭了……

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 Author| kuing Posted at 2013-12-10 00:58:01
回复 6# isee

选择-10

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