The product of all positive real values of $x$ satisfying the equation $x^{\left(16(\log_5 x)^3 - 68 \log_5 x\right)} = 5^{-16}$ is ________.

$\text{Given: } x^{\left(16(\log_5 x)^3 - 68\log_5 x\right)} = 5^{-16}, \quad x>0$ $\text{To find: } \prod x, \text{ over all positive real } x \text{ satisfying the equation.}$ $t = \log_5 x \Rightarrow x = 5^{t}$ $\left(5^{t}\right)^{16t^{3}-68t} = 5^{\,t\left(16t^{3}-68t\right)}$ $= 5^{\,16t^{4}-68t^{2}}$ $5^{\,16t^{4}-68t^{2}} = 5^{-16}$ $\Rightarrow 16t^{4}-68t^{2} = -16$ $16t^{4}-68t^{2}+16 = 0$ $\Rightarrow 4t^{4}-17t^{2}+4 = 0$ $u = t^{2} \Rightarrow 4u^{2}-17u+4 = 0$ $\Delta = 17^{2}-4\cdot4\cdot4 = 289-64 = 225, \quad \sqrt{\Delta}=15$ $u = \dfrac{17\pm15}{8}$ $u = 4 \quad \text{or} \quad u = \dfrac{1}{4}$ $t^{2}=4 \Rightarrow t=\pm2; \quad t^{2}=\dfrac{1}{4} \Rightarrow t=\pm\dfrac{1}{2}$ $x = 5^{t} \Rightarrow x \in \left\{25,\ \dfrac{1}{25},\ \sqrt{5},\ \dfrac{1}{\sqrt{5}}\right\}$ $\prod x = 5^{\sum t} = 5^{\,2-2+\frac12-\frac12} = 5^{0} = 1$ $\boxed{1}$