To find the coefficient of $x^7$ in the expansion of $(2x - 3)^7$, we can use the Binomial Theorem. The Binomial Theorem states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In this case, $a = 2x$, $b = -3$, and $n = 7$.

We want the term where the power of $x$ is exactly 7. In the expansion, the general term looks like this:

$T_k = \binom{7}{k} (2x)^{7-k} (-3)^k$

We need the power of $x$ to be 7, which means the exponent of $x$ in $(2x)^{7-k}$ must be 7. This occurs when $7 - k = 7$, or $k = 0$.

Now, let's calculate the term when $k = 0$:

$T_0 = \binom{7}{0} (2x)^7 (-3)^0$

This simplifies to:

$T_0 = 1 \cdot (2x)^7 \cdot 1 = (2x)^7 = 128x^7$

Thus, the coefficient of $x^7$ is $128$.

The correct answer is 128.

Would you like more details on this process, or do you have any questions?

Here are 5 related questions for further exploration:

1. How would the coefficient change if the exponent were higher, say $(2x - 3)^{10}$?
2. What if the signs inside the binomial were different, such as $(2x + 3)^7$?
3. How can the Binomial Theorem be applied to expansions with fractional exponents?
4. How would you expand $(2x - 3)^7$ without using the Binomial Theorem?
5. What is the general form of the Binomial Theorem for negative exponents?

Tip: Always double-check the signs in binomial expansions, especially when terms involve subtraction!