To solve the equation \((2x - 3)^4 = (2x - 3)^6\), we first set \(y = (2x - 3)\). This simplifies our equation to:

\[
y^4 = y^6
\]

Next, we can rearrange it to:

\[
y^6 - y^4 = 0
\]

Factoring out \(y^4\), we get:

\[
y^4(y^2 - 1) = 0
\]

This implies two cases:

1. \(y^4 = 0\)
2. \(y^2 - 1 = 0\)

For the first case:

\[
y^4 = 0 \implies y = 0 \implies 2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}
\]

For the second case:

\[
y^2 - 1 = 0 \implies y^2 = 1 \implies y = 1 \text{ or } y = -1
\]

If \(y = 1\):

\[
2x - 3 = 1 \implies 2x = 4 \implies x = 2
\]

If \(y = -1\):

\[
2x - 3 = -1 \implies 2x = 2 \implies x = 1
\]

Thus, the solutions are:

\[
x = \frac{3}{2}, \quad x = 2, \quad x = 1
\]

The final answer is:

\[
x = \frac{3}{2}, \quad x = 1, \quad x = 2
\]