Salut,
Logaritmăm fiecare relație din enunț în baza 3:
[tex]log_3{45^a}=log_35,\ sau\ a\cdot log_3(3^2\cdot 5)=log_35,\ sau\ a\cdot(2+log_35)=log_35;\\log_3{45^b}=log_33,\ sau\ b\cdot log_3(3^2\cdot 5)=1,\ sau\ b\cdot(2+log_35)=1;\\\\1-a-b=1-\dfrac{log_35}{2+log_35}-\dfrac{1}{2+log_35}=\dfrac{2+log_35-log_35-1}{2+log_35}=\dfrac{1}{2+log_35}\ (1)\\\\1-a=1-\dfrac{log_35}{2+log_35}=\dfrac{2+log_35-log_35}{2+log_35}=\dfrac{2}{2+log_35}\ (2).\\\\Din\ rela\c{t}iile\ (1)+(2)\ avem\;c\breve{a}:\dfrac{1-a-b}{1-a}=\dfrac{1}2\Rightarrow 16^{\frac{1}2}=\sqrt{16}=4\in\mathbb{Z}.[/tex]
Simplu, nu ? :-).
Green eyes.
