To solve \(25! = 2^x\), we need to find how many factors of 2 are in \(25!\). This can be done by using the formula for counting the number of factors of a prime \(p\) in \(n!\):

\[
\text{Number of factors of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
\]

Here, \(n = 25\) and \(p = 2\).

Let's compute the number of factors of 2 in \(25!\):

\[
\left\lfloor \frac{25}{2} \right\rfloor = 12
\]
\[
\left\lfloor \frac{25}{4} \right\rfloor = 6
\]
\[
\left\lfloor \frac{25}{8} \right\rfloor = 3
\]
\[
\left\lfloor \frac{25}{16} \right\rfloor = 1
\]

Now, sum them up:

\[
12 + 6 + 3 + 1 = 22
\]

So, the number of factors of 2 in \(25!\) is 22. Therefore, \(x = 22\).

### Final Answer: 
\[
x = 22
\]

Would you like more details or further explanation on this? Here are some related questions:

1. How would you calculate the number of factors of 3 in \(25!\)?
2. Can this method be applied to any factorial?
3. How is this method derived from prime factorization?
4. What would happen if the factorial was larger, say \(50!\)?
5. How does \(2^x = n\) change if \(n\) includes factors other than 2?

**Tip**: When finding powers of primes in factorials, this method works for any prime, not just 2!

Learn how to solve the equation 25! = 2^x by using prime factorization to find the number of factors of 2 in 25!. This step-by-step solution is suitable for students in Grades 9-12 and covers important mathematical concepts like factorials, prime factorization, and exponents.

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