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Right to left exponentiation https://www.calcudoku.org/forum/viewtopic.php?f=2&t=657 
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Author:  sjs34 [ Mon Jan 19, 2015 1:30 pm ] 
Post subject:  Right to left exponentiation 
In today's 7x7 difficult puzzle I did not understand why (2 ^ 2) ^ 5 was not a valid answer for 1024. Patrick tells me that it violates the right to left convention. I'm not sure that I understand that, but at least the autocorrect will tell me when I'm in the wrong zone. 
Author:  pnm [ Mon Jan 19, 2015 2:54 pm ] 
Post subject:  Re: Right to left exponentiation 
This issue pops up every once in a while, see for example: viewtopic.php?f=16&t=183&p=1651#p1651 viewtopic.php?f=2&t=49&p=389#p389 viewtopic.php?f=3&t=55 etc. 
Author:  eclipsegirl [ Mon Jan 19, 2015 5:28 pm ] 
Post subject:  Re: Right to left exponentiation 
I do not know if this will help. It is how I think about it. This issue usually arises when the solutions could be a power (multiple) of 2, 4 and 8 In a cage that takes three elements, the final exponent MUST be a power (multiple) of another number In todays example, we had the solution of 1024 which is either 2^10 or 4^5 The exponent 10 in not a power of any other digit, so that solution can not work That leaves 4^5 as the only solution. 5 can only be 5^1. The solution set for cage is {1, 4, 5} There was a puzzle recently that had 65536 has the solution. 63356 = 2^16 or 4^ 8 It was also a bent cage of thee elements. The possible solution sets were {2, 2, 4} for 2 ^ (4^2) = 2^16 or {2, 3, 4} for 4 ^ (2^3) = 4^8 So until elimination from other rows forced a solution, one had to keep the numbers (2,3,4) in all three positions of the cage. I think the correct answer took advantage of the bent (corner) cage and was {2, 4, 2} with the 4 being in the corner of the bent cage 
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