Now given that $e^y = x-r$ , $r \in \mathbb{Z}^+$ has one real root only and state the range of $r$.
For a start $y = \ln(x-r)$ and hence the root of it should be $x = r+1$. How should I deduce the answer $r \geq 2?$
2026-03-31 09:57:30.1774951050
$e^y = x-r$ , $r \in \mathbb{Z}^+$ has one real root only and state the range of $r$.
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Since $r\in\mathbb Z^+=\{1,2,3,\cdots\}$, $$x=(r+1)\in\{2,3,4,\cdots\}$$
If the question is asking for what value of $r$ does the equation $e^y=x-r$ has one unique real root, then clearly the answer is all $r\in\mathbb R$.