What is the sum of real roots of $(x\sqrt{x})^x = x^{x\sqrt{x}}$?

$ \textbf{(A)}\ \frac{18}{7}
\qquad\textbf{(B)}\ \frac{71}{4}
\qquad\textbf{(C)}\ \frac{9}{4}
\qquad\textbf{(D)}\ \frac{24}{19}
\qquad\textbf{(E)}\ \frac{13}{4}
$
1. Start with the given equation:
   \[
   (x\sqrt{x})^x = x^{x\sqrt{x}}
   \]

2. Simplify the left-hand side:
   \[
   (x\sqrt{x})^x = (x^{3/2})^x = x^{x \cdot \frac{3}{2}} = x^{\frac{3x}{2}}
   \]

3. Now, the equation becomes:
   \[
   x^{\frac{3x}{2}} = x^{x\sqrt{x}}
   \]

4. Since the bases are the same, we can equate the exponents:
   \[
   \frac{3x}{2} = x\sqrt{x}
   \]

5. Divide both sides by \(x\) (assuming \(x \neq 0\)):
   \[
   \frac{3}{2} = \sqrt{x}
   \]

6. Square both sides to solve for \(x\):
   \[
   \left(\frac{3}{2}\right)^2 = x
   \]
   \[
   \frac{9}{4} = x
   \]

7. Verify if there are any other solutions. Consider \(x = 1\):
   \[
   (1\sqrt{1})^1 = 1^{1\sqrt{1}} \implies 1 = 1
   \]
   This is true, so \(x = 1\) is also a solution.

8. The real roots are \(x = \frac{9}{4}\) and \(x = 1\).

9. Sum the real roots:
   \[
   \frac{9}{4} + 1 = \frac{9}{4} + \frac{4}{4} = \frac{13}{4}
   \]

The final answer is \(\boxed{\frac{13}{4}}\).