Determine all possible real numbers x that satisfy this equation:

2(2^x- 1)x² + (2^{x^2}- 2)x = 2^x+1 - 2

Prove that there are no others.

First by inspection there are solutions x={-1,0,1}

Second rewrite the equation a bit
2(2^x- 1)x² + (2^{x^2}- 2)x = 2^x+1 - 2
2(2^x- 1)x² + (2^{x^2}- 2)x = 2(2^x - 1)
2(2^x- 1)x² - 2(2^x - 1)= -(2^{x^2}- 2)x
2(2^x- 1)(x² -1) = -(2^{x^2}- 2)x
2(2^x- 1)/(2^{x^2}- 2) = -x/(x² -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)
(x² -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 2015-12-06 17:29:16