E.444. Determinați numărul natural n,n,n, astfel încât 33n+1+2⋅33n+327<329−328.3^{3^n+1} + 2 \cdot 3^{3^n} + 3^{27} < 3^{29}-3^{28}.33n+1+2⋅33n+327<329−328.
Răspuns: n∈{0,1,2}.n \in \{0,1,2\}.n∈{0,1,2}.
33n+1+2⋅33n<329−328−3273^{3^n+1} + 2 \cdot 3^{3^n} < 3^{29}-3^{28} - 3^{27}33n+1+2⋅33n<329−328−327 33n(3+2)<327(32−3−1)∣:53^{3^n}(3+2) < 3^{27}(3^2 - 3 - 1) \quad |:533n(3+2)<327(32−3−1)∣:5 33n<3273^{3^n} < 3^{27}33n<327 ⇒3n<27⇒n∈{0,1,2}.\Rightarrow 3^n < 27 \Rightarrow \boxed{n \in \{0,1,2\}}.⇒3n<27⇒n∈{0,1,2}.