SOLUTION: Find the integer k, k > 2, for, which log (k - 2)! + log(k - 1)! + 2 = 2 log k!.
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Question 530282: Find the integer k, k > 2, for, which log (k - 2)! + log(k - 1)! + 2 = 2 log k!.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
I presume that the factorial is inside the log, otherwise it would not easily be defined.
We can write the original equation,
!) + \log((k-1)!) + 2 = 2\log(k!))
, as
!(k-1)!) + 2 = \log((k!)^2))
, using logarithmic properties.
^2) - \log((k-2)!(k-1)!) = 2)
^2}{(k-2)!(k-1)!}) = 2)
(k-1)) = 2)
, upon simplifying the factorial expression in the LHS. Assuming the log is in base 10,
(k-1) = 100)
, in which k = 5 is the unique positive integer solution.
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