Question

comparati nr a=9 la puterea 15 si b=27 la puerea 11

Answer 1

$9^{15}=(3^2)^{15}<=>3^{2*15}<=>3^{30} \\ \\ 27^{11}=(3^3)^{11}<=>3^{3*11}<=>3^{33} \\ \\ 3^{30}<3^{33}<=>\boxed{\boxed{9^{15}<27^{11}}}$

Answer 2

$9^{15}=3^{2^{15}}=3^{30}$
$27^{11}=3^{3^{11}}=3^{33}$ 
    De aici tragem concluzia ca: a<b