The sum of the exponents

There is only $1$ possible value of $x$ , that is $0$ .

Sum of real numbers satisfying $x$ in the equation = $\boxed{0}$

Let $a=2^x$ , $b=3^x$ , $c=1^x$ . We have

$\begin{aligned} ac + bc - a^2 + ab - b^2 &= c^2\\ a^2 + b^2 + c^2 - ab - bc - ca &= 0\\ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca &= 0\\ (a-b)^2 + (b-c)^2 + (c-a)^2 &=0\\ \end{aligned}$

Since squares are non-negative, we must have each of the three squares equal to 0. Thus, $a=b=c$ so $2^x = 3^x = 1^x = 1$ so $x=0$ .

Split the terms as : $4^{x} = 2^{x} * 2^{x}$ , $9^{x}$ as $3^{x} * 3^{x}$ , $6^{x}$ as $2^{x}* 3^{x}.$ , and so on .. Take $3^{x}=a$ , $2^{x}= b$ , $1^{x} =c ,$ then $a^{2} + b^{2} + c^{2}$ = ab+ bc + ca . The eqn. has one sol. with a= b= c= 1. Hence $2^{x} = 3^{x} =1$ So. $x=\boxed{0}.$