Take the base-5 logarithm on both sides and simplify.
[tex]x^{16 \log^3_5(x) - 68 \log_5(x)} = 5^{-16}[/tex]
[tex]\left(16 \log^3_5(x) - 68 \log_5(x)\right) \log_5(x) = -16[/tex]
[tex]16 \log^4_5(x) - 68 \log^2_5(x) + 16 = 0[/tex]
[tex]4 \log^4_5(x) - 17 \log^2_5(x) + 4 = 0[/tex]
Factorize the left side.
[tex]\left(4 \log^2_5(x) - 1\right) \left(\log^2_5(x) - 4\right) = 0[/tex]
Then
[tex]4 \log^2_5(x) - 1 = 0 \implies \log^2_5(x) = \dfrac14 \\\\ ~~~~~~~~ \implies \log_5(x) = \pm \dfrac12 \\\\ ~~~~~~~~ \implies x = \dfrac1{\sqrt5} \text{ or } x = \sqrt5[/tex]
or
[tex]\log^2_5(x) - 4 = 0 \implies \log^2_5(x) = 4 \\\\ ~~~~~~~~ \implies \log_5(x) = \pm 2 \\\\ ~~~~~~~~ \implies x = \dfrac1{25} \text{ or } x = 25[/tex]
and the product of these solutions is of course 1.