Problem Statement:
Given f'(1)=1 and f(2x)=f(x) for x>0. If f'(x) is differentiable, then there exists a number c in the interval (2,4) such that f"(c) equals what?

Approach:
To find the value of f"(c), we will use the given information and apply the Mean Value Theorem (MVT) in two steps. First, we will find the average rate of change of f'(x) over the interval [1,2], and then we will apply MVT to the function f'(x) on the interval [1,2].

Solution:

Step 1: Average Rate of Change
Let's find the average rate of change of f'(x) over the interval [1,2]. We will use the formula:
\[ \text{{Average Rate of Change}} = \frac{{f'(2) - f'(1)}}{{2 - 1}} \]

Given f'(1) = 1 and f(2x) = f(x), we can differentiate f(2x) with respect to x using the Chain Rule:
\[ f'(2x) \cdot 2 = f'(x) \cdot 1 \]
\[ f'(2x) = \frac{{f'(x)}}{2} \]

Now we can substitute f'(2x) into the average rate of change formula:
\[ \text{{Average Rate of Change}} = \frac{{\frac{{f'(x)}}{2} - 1}}{1} \]
\[ \text{{Average Rate of Change}} = \frac{{f'(x) - 2}}{{2}} \]

Step 2: Applying the Mean Value Theorem (MVT)
According to the Mean Value Theorem, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the interval (a, b) such that:
\[ f'(c) = \frac{{f(b) - f(a)}}{{b - a}} \]

In our case, we will apply MVT to the function f'(x) on the interval [1,2]:
\[ f'(c) = \frac{{f'(2) - f'(1)}}{{2 - 1}} \]
\[ f'(c) = \frac{{\frac{{f'(x)}}{2} - 1}}{1} \]
\[ f'(c) = \frac{{f'(x) - 2}}{{2}} \]

Comparing the results from Step 1 and Step 2, we can see that the average rate of change of f'(x) over the interval [1,2] is equal to f'(c) as per MVT.

Conclusion:
Therefore, there exists a number c in the interval (2,4) such that f"(c) is equal to the average rate of change of f'(x) over the interval [1,2], which is (f'(x) - 2)/2.