First by inspection there are solutions x={1,0,1}
Second rewrite the equation a bit
2(2
^{x} 1)x
^{2} + (2
^{x^2} 2)x = 2
^{x+1}  2
2(2
^{x} 1)x
^{2} + (2
^{x^2} 2)x = 2(2
^{x}  1)
2(2
^{x} 1)x
^{2}  2(2
^{x}  1)= (2
^{x^2} 2)x
2(2
^{x} 1)(x
^{2} 1) = (2
^{x^2} 2)x
2(2
^{x} 1)/(2
^{x^2} 2) = x/(x
^{2} 1)
Third compare each by intervals
(2
^{x} 1) is  on (inf,0) and + on (0,inf)
(2
^{x^2} 2) is  on (1,1) and  on (inf,1), (1,inf)
x is + on (inf,0) and  on (0,inf)
(x
^{2} 1) is  on (1,1) and + on (inf,1), (1,inf)
Finally, since these quotients are each + where the other is  they can never be equal. So there are no other solutions.

Posted by Jer
on 20151206 17:29:16 