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

Let $$y = \log_5 x$$. Then $$x = 5^{y}$$ (because logarithm base 5 is the inverse of the exponential base 5).

Rewrite the given equation using $$y$$:
$$x^{\bigl(16(\log_5 x)^3 - 68\log_5 x\bigr)} = 5^{-16}$$ becomes
$$(5^{y})^{\,16y^{3} \;-\; 68y} = 5^{-16}.$$

Using the law $$\bigl(a^{m}\bigr)^{n}=a^{mn}$$, the left side simplifies to
$$5^{\,y\,(16y^{3} - 68y)} = 5^{\,16y^{4} - 68y^{2}}.$$

Because the bases are equal (both are 5), equate the exponents:
$$16y^{4} - 68y^{2} = -16.$$

Divide every term by 2 to make the numbers smaller:
$$8y^{4} - 34y^{2} + 8 = 0 \qquad -(1)$$

Put $$z = y^{2} \;(\text{note: } z \ge 0)$$ to convert the quartic to a quadratic:

Equation $$(1)$$ becomes
$$8z^{2} - 34z + 8 = 0.$$

Solve this quadratic using the quadratic formula $$z = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ with $$a=8,\, b=-34,\, c=8$$:

Discriminant $$\Delta = (-34)^{2} - 4\cdot 8 \cdot 8 = 1156 - 256 = 900.$$
$$\sqrt{\Delta} = 30.$$

Therefore
$$z = \dfrac{34 \pm 30}{16}.$$

Case 1: $$z = \dfrac{34 + 30}{16} = \dfrac{64}{16} = 4.$$
Case 2: $$z = \dfrac{34 - 30}{16} = \dfrac{4}{16} = \dfrac14.$$

Recall $$z = y^{2}$$, so

From Case 1: $$y^{2}=4 \quad\Longrightarrow\quad y = \pm 2.$$
From Case 2: $$y^{2}=\dfrac14 \quad\Longrightarrow\quad y = \pm \dfrac12.$$

Convert back to $$x$$ using $$x = 5^{y}$$:

For $$y = 2: \; x = 5^{2} = 25.$$
For $$y = -2: \; x = 5^{-2} = \dfrac1{25}.$$
For $$y = \dfrac12: \; x = 5^{1/2} = \sqrt{5}.$$
For $$y = -\dfrac12: \; x = 5^{-1/2} = \dfrac1{\sqrt{5}}.$$

All four values are positive real numbers, so they all satisfy the original equation.

The required product is
$$25 \times \dfrac1{25} \times \sqrt{5} \times \dfrac1{\sqrt{5}} = 1.$$

Hence, the product of all positive real solutions is 1.