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June 1st, 2019, 04:20 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81  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: 528 Thanks: 42 
$n = 1, 2$ That seems too easy. Whatâ€™s the catch? 
June 2nd, 2019, 03:51 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81  n must be power of 2 which gives $\displaystyle 2^k =k+1 $ . k=0 and 1 . 
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 Last edited by gedmathbootcamp; June 2nd, 2019 at 01:26 PM. 
June 2nd, 2019, 06:23 PM  #5 
Senior Member Joined: Aug 2012 Posts: 2,321 Thanks: 714  

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