Răspuns:
Explicație pas cu pas:
[tex]2^{a},~3^{a+1},~2^{a+2}~termeni~consecutivi~a~unei~progresii~aritmetice,~deci\\3^{a+1}=\dfrac{2^{a}+2^{a+2}}{2},~=>~3*3^{a}=\dfrac{2^{a}+2^{a}*2^{2}}{2} ,~=>~6*3^{a}=2^{a}*(1+2^{2}),~=>~ 6*3^{a}=5*2^{a}~|:(6*2^{a}) ,~=>~\dfrac{6*3^{a}}{6*2^{a}} =\dfrac{5*2^{a}}{6*2^{a}},~=>~(\frac{3}{2})^{a}=\frac{5}{6},~=>~a=log_{\frac{3}{2}}\frac{5}{6}.[/tex]Rezolvare2
[tex]2^{a},~3^{a}+1,~2^{a+2}\\2*(3^{a}+1)=2^{a}+2^{a}*2^{2},~=>~2*(3^{a}+1)=2^{a}*(1+2^{2}),~\\=>2*(3^{a}+1)=2^{a}*5[/tex]
Pentru a=2, obținem egalitate adevărată.
2·(3²+1)=2²·5, deci 2·10=4·5
Răspuns: a=2.
