Question 1209853
Let's solve the equation step by step.

Given equation:
$$\sqrt[5]{7} x^{\log_7 x} = x^{\log_5 x}$$

Rewrite $\sqrt[5]{7}$ as $7^{1/5}$:
$$7^{1/5} x^{\log_7 x} = x^{\log_5 x}$$

Take the logarithm base 7 of both sides:
$$\log_7 (7^{1/5} x^{\log_7 x}) = \log_7 (x^{\log_5 x})$$

Using logarithm properties:
$$\log_7 (7^{1/5}) + \log_7 (x^{\log_7 x}) = \log_5 x \cdot \log_7 x$$
$$\frac{1}{5} \log_7 7 + \log_7 x \cdot \log_7 x = \log_5 x \cdot \log_7 x$$
$$\frac{1}{5} + (\log_7 x)^2 = \log_5 x \cdot \log_7 x$$

Use the change of base formula $\log_5 x = \frac{\log_7 x}{\log_7 5}$:
$$\frac{1}{5} + (\log_7 x)^2 = \frac{\log_7 x}{\log_7 5} \cdot \log_7 x$$
$$\frac{1}{5} + (\log_7 x)^2 = \frac{(\log_7 x)^2}{\log_7 5}$$

Let $y = \log_7 x$:
$$\frac{1}{5} + y^2 = \frac{y^2}{\log_7 5}$$
$$\frac{1}{5} = \frac{y^2}{\log_7 5} - y^2$$
$$\frac{1}{5} = y^2 \left( \frac{1}{\log_7 5} - 1 \right)$$
$$y^2 = \frac{1}{5} \cdot \frac{1}{\frac{1}{\log_7 5} - 1}$$
$$y^2 = \frac{1}{5} \cdot \frac{\log_7 5}{1 - \log_7 5}$$

Since $y = \log_7 x$, we have:
$$(\log_7 x)^2 = \frac{\log_7 5}{5(1 - \log_7 5)}$$

Taking the square root:
$$\log_7 x = \pm \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}}$$

Let $a$ and $b$ be the two positive roots. Then:
$$\log_7 a = \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}}$$
$$\log_7 b = -\sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}}$$

We want to find $ab$. We know that $\log_7 (ab) = \log_7 a + \log_7 b$:
$$\log_7 (ab) = \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}} - \sqrt{\frac{\log_7 5}{5(1 - \log_7 5)}} = 0$$

Since $\log_7 (ab) = 0$, we have $ab = 7^0 = 1$.

Therefore, $ab = 1$.

Final Answer: The final answer is $\boxed{1}$