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What is the value of n such that 81^n = 9^(27) * 27^9?

$$81^n~~ = ~~9^{27}\cdot 27^9\qquad \begin{cases} 81=3\cdot 3\cdot 3\cdot 3\\ \qquad 3^4\\ 9=3\cdot 3\\ \qquad 3^2\\ 27=3\cdot 3\cdot 3\\ \qquad 3^3 \end{cases}\implies (3^4)^n~~ = ~~(3^2)^{27}\cdot (3^3)^9 \\\\\\ 3^{4n}~~ = ~~3^{2\cdot 27}\cdot 3^{3\cdot 9}\implies 3^{4n}~~ = ~~3^{54}\cdot 3^{27}\implies 3^{4n}~~ = ~~3^{54+27} \\\\\\ 3^{4n}~~ = ~~3^{81}\implies 4n=81\implies n=\cfrac{81}{4}\implies n=20\frac{1}{4}$$