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 June 1st, 2019, 04:20 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 Solve for integers Solve the equation for $\displaystyle n\in \mathbb{N} ;$. $\displaystyle 2n=2^n$.
 June 1st, 2019, 06:52 AM #2 Senior Member   Joined: Jun 2014 From: USA Posts: 564 Thanks: 43 $n = 1, 2$ That seems too easy. What’s the catch? Thanks from topsquark
 June 2nd, 2019, 03:51 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 n must be power of 2 which gives $\displaystyle 2^k =k+1$ . k=0 and 1 . Thanks from topsquark
 June 2nd, 2019, 01:17 PM #4 Newbie   Joined: Jun 2019 From: US Posts: 1 Thanks: 1 The answers are n = 1 and n = 2. n=0 doesn't work because n is not a natural number, also if n is 0 we get 0 = 1 which is not true. The question says that N must be a natural number, that means 1,2,3,4,..... The strategy here is to start trying beginning with the smallest n. Try 1, it works, try 2, it also works. If we try 3, we get 6 = 8, that doesn't work. Try 4 we get 4 = 16, that also doesn't work. We can see that the right side grows faster than the left as n increases, so no number higher than 2 will work. The only solutions are 1 and 2 Thanks from topsquark Last edited by gedmathbootcamp; June 2nd, 2019 at 01:26 PM.
June 2nd, 2019, 06:23 PM   #5
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Quote:
 Originally Posted by gedmathbootcamp The question says that N must be a natural number, that means 1,2,3,4,.....
$0$ is not a solution of $\displaystyle 2n=2^n$ regardless of which convention you use for whether $0 \in \mathbb N$. The left hand side is $0$ and the right hand side is $1$.

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