Răspuns:
[tex] {2}^{15} \times {4}^{20} \times {8}^{3} = {2}^{15} \times ({2}^{2} ) ^{20} \times ( {2}^{3} ) ^{3} = {2}^{15} \times {2}^{40} \times {2}^{9} = {2}^{15 + 40 + 9} = {2}^{64 } \\ ({2}^{8} ) ^{2} = {2}^{8 \times 2} = {2}^{16} \\ {2}^{64} > {2}^{16} \: deoarece \: 64 > 16[/tex]