Question 530282
I presume that the factorial is inside the log, otherwise it would not easily be defined.


We can write the original equation,


*[tex \LARGE \log((k-2)!) + \log((k-1)!) + 2 = 2\log(k!)], as


*[tex \LARGE \log((k-2)!(k-1)!) + 2 = \log((k!)^2)], using logarithmic properties. 


*[tex \LARGE \log((k!)^2) - \log((k-2)!(k-1)!) = 2]


*[tex \LARGE \log(\frac{(k!)^2}{(k-2)!(k-1)!}) = 2]


*[tex \LARGE \log(k(k)(k-1)) = 2], upon simplifying the factorial expression in the LHS. Assuming the log is in base 10,


*[tex \LARGE k(k)(k-1) = 100], in which k = 5 is the unique positive integer solution.