Find the positive integer $n$ such that $32$ is the product of the real number solutions of $x^{\log_2(x^3)-n} = 13$
1. Start with the given equation:
   \[
   x^{\log_2(x^3) - n} = 13
   \]

2. Take the logarithm base \( x \) of both sides:
   \[
   \log_x \left( x^{\log_2(x^3) - n} \right) = \log_x 13
   \]

3. Simplify the left-hand side using the property of logarithms \(\log_b(a^c) = c \log_b(a)\):
   \[
   \log_2(x^3) - n = \log_x 13
   \]

4. Express \(\log_2(x^3)\) in terms of \(\log_2 x\):
   \[
   \log_2(x^3) = 3 \log_2 x
   \]

5. Substitute this into the equation:
   \[
   3 \log_2 x - n = \log_x 13
   \]

6. Convert \(\log_x 13\) to base 2 using the change of base formula \(\log_b a = \frac{\log_c a}{\log_c b}\):
   \[
   \log_x 13 = \frac{\log_2 13}{\log_2 x}
   \]

7. Substitute this into the equation:
   \[
   3 \log_2 x - n = \frac{\log_2 13}{\log_2 x}
   \]

8. Multiply both sides by \(\log_2 x\) to clear the fraction:
   \[
   (3 \log_2 x - n) \log_2 x = \log_2 13
   \]

9. Let \( y = \log_2 x \). Then the equation becomes:
   \[
   3y^2 - ny = \log_2 13
   \]

10. Rearrange this into a standard quadratic equation form:
    \[
    3y^2 - ny - \log_2 13 = 0
    \]

11. Solve the quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
    \[
    y = \frac{n \pm \sqrt{n^2 + 12 \log_2 13}}{6}
    \]

12. Since \( y = \log_2 x \), the solutions for \( x \) are:
    \[
    x = 2^{\frac{n \pm \sqrt{n^2 + 12 \log_2 13}}{6}}
    \]

13. The product of the two solutions for \( x \) is given as \( 2^5 \):
    \[
    2^{\frac{n + \sqrt{n^2 + 12 \log_2 13}}{6}} \cdot 2^{\frac{n - \sqrt{n^2 + 12 \log_2 13}}{6}} = 2^5
    \]

14. Simplify the product:
    \[
    2^{\frac{n + \sqrt{n^2 + 12 \log_2 13} + n - \sqrt{n^2 + 12 \log_2 13}}{6}} = 2^5
    \]
    \[
    2^{\frac{2n}{6}} = 2^5
    \]
    \[
    2^{\frac{n}{3}} = 2^5
    \]

15. Equate the exponents:
    \[
    \frac{n}{3} = 5
    \]
    \[
    n = 15
    \]

The final answer is \( \boxed{15} \).