[tex]\frac{2^x + a \times 3^x}{2^x - a\times 3^x}=2\\2^x+a\times 3^x = 2(2^x - a\times 3^x)\\2^x + a\times 3^x = 2\times2^x-2a \times 3^x \\ 2^x = 3a\times 3^x\\2^x = a\times 3^{x+1}\\x = log_2 (2^x) = log_2 (a \times 3^{x+1})=log_2(a) + log_2(3^{x+1})=log_2 (a) + (x+1)\times log_2(3)\\log_2(a)= x - (x+1)\times log_2 (3)\\log_2 (a) = x - xlog_2(3) - log_2(3) = x(1 - log_2(3)) - log_2(3)\\log_2(a) + log_2(3) = x(1 - log_2(3))\\x = \frac{log_2(a)+log_2(3)}{1 - log_2(3)}=\frac{log_2(3a)}{log_2(2)-log_2(3)}=\frac{log_2(3a)}{log_2(2/3)}=log_{2/3}(3a)[/tex]
Atunci x este intreg numai daca [tex]3a = (2/3)^k[/tex], cu k din Z
[tex]3a = \frac{2^k}{3^k}\\a=\frac{2^k}{3^{k+1}}[/tex]
