Bài 1. Ta có : \({3^2}.{3^4}.{\rm{ }}{3^{n}} = {3^{2 + 4 + n}} \Rightarrow {3^{6 + n}} = {3^{10}}\)
\(6+ n = 10\Rightarrow n = 10 – 6 = 4.\)
Bài 2. Ta có : \(9 = {3^2};243 = {3^{5}} \Rightarrow {3^2} < {3^x} \le {\rm{ }}{3^5}\) \(\Rightarrow x ∈ \{3; 4 ; 5 \}\). Vậy \(A = \{3; 4; 5 \}.\)
Bài 3. Ta có :
\(\eqalign{ & {2^{30}} = \underbrace {{2^3}{{.2}^3}{{.....2}^3}}_{10} = {8^{10}}; \cr & {3^{20}} = \underbrace {{3^2}{{.3}^2}{{....3}^2}}_{10} = {9^{10}} \cr} \)
Vì \(8 < 9 \Rightarrow {8^{10}} < {9^{10}} \Rightarrow {2^{30}} < {3^{20}}\)
