Odpowiedź:
[tex]{3}^{2^2}<(3^{2})^3 <3^{3^2} <(27^2)^2<{27}^{2^3}[/tex]
Szczegółowe wyjaśnienie:
[tex](27^2)^2 = ((3^3)^2)^2=3^{3*2*2} =3^{12}\\{3}^{2^2} = 3^{4}\\{27}^{2^3} = {(3^3)^8}} = 3^{3*8} = 3^{24}\\3^{3^2} = 3^9 \\(3^{2})^3 = 3^{2*3} = 3^6\\\\3^4 < 3^6 <3^9<3^{12}<3^{24}\\czyli\\{3}^{2^2}<(3^{2})^3 <3^{3^2} <(27^2)^2<{27}^{2^3}[/tex]
