Product of Real Values Satisfying Logarithmic Exponent Equation

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

we take the logarithm with base $5$ on both sides of the equation: $\log_5 \left( x^{\left(16(\log_5 x)^3 - 68 \log_5 x\right)} \right) = \log_5 \left(5^{-16}\right)$

Using the logarithmic power property $\log_b(a^c) = c \log_b a$, we get: $\left(16(\log_5 x)^3 - 68 \log_5 x\right) \log_5 x = -16$

Let $y = \log_5 x$. Substituting $y$ into the equation gives: $\left(16 y^3 - 68 y\right) y = -16$ $16 y^4 - 68 y^2 + 16 = 0$

Dividing the entire equation by $4$: $4 y^4 - 17 y^2 + 4 = 0$

Let $z = y^2$. The equation becomes a quadratic equation in terms of $z$: $4 z^2 - 17 z + 4 = 0$

Factoring the quadratic equation: $4 z^2 - 16 z - z + 4 = 0$ $4z(z - 4) - 1(z - 4) = 0$ $(4z - 1)(z - 4) = 0$

Thus, the roots for $z$ are: $z = 4 \quad \text{or} \quad z = \frac{1}{4}$

Since $z = y^2$, we solve for $y$:

1. $y^2 = 4 \implies y = 2 \quad \text{or} \quad y = -2$
2. $y^2 = \frac{1}{4} \implies y = \frac{1}{2} \quad \text{or} \quad y = -\frac{1}{2}$

The four real solutions for $y = \log_5 x$ are $y_1 = 2$, $y_2 = -2$, $y_3 = \frac{1}{2}$, and $y_4 = -\frac{1}{2}$.

The corresponding positive values of $x$ are given by $x = 5^y$: $x_1 = 5^2, \quad x_2 = 5^{-2}, \quad x_3 = 5^{1/2}, \quad x_4 = 5^{-1/2}$

The product $P$ of all these positive real values of $x$ is: $P = x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 5^{y_1 + y_2 + y_3 + y_4}$ $P = 5^{2 + (-2) + \frac{1}{2} + \left(-\frac{1}{2}\right)} = 5^0 = 1$

Thus, the product of all positive real values of $x$ satisfying the equation is $1$.