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Let log denote log base $3$. Then $a=log(n)+log(n-1)$.
Find the sum of all $n$ such that $n^{\log _{3} {n-1}}+2(n-1)^{\log _{3} {n}}=3n^2$.
Let log denote log base $3$. Then $a=log(n)+log(n-1)$.