a)
·[tex]21^{22} < 5^{33} 2^{22} \\21^{22} < 5^{22} 5^{11} 2^{22}[/tex]
[tex](\frac{21}{5} )^{22} < 5^{11} 2^{22}[/tex]
[tex](\frac{21}{10} )^{22} < 5^{11}[/tex]
[tex](\frac{21}{10} )^{11} (\frac{21}{10} )^{11} <5^{11}[/tex]
Imparti relatia la 5^11 , rezultand :
[tex](\frac{21}{10} )^{11} (\frac{21}{10} )^{11} \frac{1}{5} ^{11} < 1[/tex]
[tex](\frac{21}{10} )^{11} (\frac{21}{50} )^{11} < 1[/tex]
[tex](\frac{441 }{500} )^{11} < 1\\[/tex]
[tex]\frac{441}{500} < 1\\ 441 < 500[/tex]
b) Demonstram prin absurd ca a < b .
[tex]2^{23} 5^{35} <3^{22} 7^{24}[/tex]
[tex]2^{22} *2*5^{22} *5^{13} < 3^{22} * 7^{22} * 7^{2}[/tex]
[tex](\frac{2*5}{3*7} )^{22} * 2 < 7^{2}[/tex]
[tex](\frac{2*5}{3*7} )^{22} < \frac{ 49 }{2}[/tex]
[tex](\frac{10}{21} )^{22} < \frac{ 49 }{2}[/tex]
Cum 10 / 21 < 1 inseamna ca orice numar subunitar ridicat la o putere tinde sa se micsoreze(tinde spre zero).
Astfel relatia [tex](\frac{10}{21} )^{22} < \frac{ 49 }{2}[/tex] e adevarata => a < b
