Sejam $a = 2^x$ e $b = 3^x$, do enunciado, temos:$$\frac{a^4b^2 + a^2b^4}{a^6 + b^6} = \frac{6}{7} \implies \frac{a^2b^2\cancel{(a^2 + b^2)}}{\cancel{(a^2 + b^2)}(a^4 - a^2b^2 + b^4)} = \frac{6}{7}$$ Temos $7a^2b^2 = 6\cdot(a^4 - a^2b^2 + b^4)$ $\implies$ $a^2b^2 = 6\cdot (a^2 - b^2)^2$ , logo: $|ab| =| \sqrt{6\space}\cdot (a^2-b^2)|$ $\implies$ $|6^x| = |\sqrt{6\space}\cdot (4^x-9^x)|$. Dividindo a equação por $6^x$, temos: $1 = \sqrt{6\space}\cdot |\left(\frac{2}{3} \right)^x-\left(\frac{3}{2} \right)^x|$ , seja $y = \left(\frac{2}{3} \right)^x$ , temos: $$\frac{1}{\sqrt 6} = y - \frac{1}{y} \implies y^2 - \frac{y}{\sqrt 6} = 1 \implies \left(y - \frac{1}{\sqrt {24}} \right)^2 = \frac{25}{24}$$ Assim: $0<y = \large{\left(\frac{2}{3} \right)^x} = \frac{6}{\sqrt{24}} = \sqrt{\frac{3}{2}}$ $\implies$ $\large{\left(\frac{2}{3} \right)^x} = \left( \frac{2}{3} \right)^{-\frac{1}{2}}$ $\implies$ $\boxed{x = -\frac{1}{2}}$ Ou: $$\frac{1}{\sqrt 6} = \frac{1}{y} - y \implies y^2 + \frac{y}{\sqrt 6} = 1 \implies \left(y + \frac{1}{\sqrt {24}} \right)^2 = \frac{25}{24}$$ Assim: $0<y = \large{\left(\frac{2}{3} \right)^x} = \frac{4}{\sqrt{24}} = \sqrt{\frac{2}{3}}$ $\implies$ $\large{\left(\frac{2}{3} \right)^x} = \left( \frac{2}{3} \right)^{\frac{1}{2}}$ $\implies$ $\boxed{x = \frac{1}{2}}$ Portanto, a soma dos módulos das soluções é: $\boxed{\left|-\frac{1}{2} \right| + \left|\frac{1}{2} \right| = 1}$