We are given the following problem involving a regular polygon:

1. The polygon has $n$ sides.
2. Each interior angle is $2x^\circ$.
3. Each exterior angle is $\left(\frac{x - 54}{2}\right)^\circ$.

Step 1: Relationship between interior and exterior angles

For any polygon, the interior and exterior angles are supplementary, meaning: $\text{Interior angle} + \text{Exterior angle} = 180^\circ$ Substituting the given expressions: $2x + \left(\frac{x - 54}{2}\right) = 180$

Step 2: Solve for $x$

First, clear the fraction by multiplying the entire equation by 2: $2(2x) + \left(x - 54\right) = 360$ Simplifying: $4x + x - 54 = 360$ $5x - 54 = 360$ Now, solve for $x$: $5x = 360 + 54 = 414$ $x = \frac{414}{5} = 82.8^\circ$

Step 3: Find the exterior angle

Using the value of $x = 82.8^\circ$, substitute it into the expression for the exterior angle: $\text{Exterior angle} = \frac{82.8 - 54}{2} = \frac{28.8}{2} = 14.4^\circ$

Step 4: Use the exterior angle to find $n$

For any regular polygon, the sum of the exterior angles is always $360^\circ$, and each exterior angle is $\frac{360^\circ}{n}$. Therefore, we have: $\frac{360}{n} = 14.4$ Solving for $n$: $n = \frac{360}{14.4} = 25$

Thus, the polygon has 25 sides.

Would you like any further clarifications on this problem? Here are some related questions you could explore:

1. How do the interior and exterior angles of a regular polygon relate?
2. What is the formula for calculating the sum of the interior angles of a polygon?
3. How would the angles change if the number of sides increased?
4. What happens to the interior angle as the number of sides approaches infinity?
5. Can the exterior angle ever be greater than the interior angle?

Tip: Remember, the exterior angle formula $\frac{360^\circ}{n}$ always holds for regular polygons, regardless of their size.