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We are asked to verify whether Rolle's Theorem applies to the given function and, if it does, to find the value of \( c \) for the function:

\[
f(x) = \log \left( \frac{x^2 + ab}{x(a + b)} \right)
\]
on the interval \([a, b]\), where \( a > 0 \).

### Rolle's Theorem Conditions:
Rolle's Theorem states that if a function \( f(x) \) satisfies the following conditions on the interval \([a, b]\):
1. **Continuity**: \( f(x) \) is continuous on the closed interval \([a, b]\).
2. **Differentiability**: \( f(x) \) is differentiable on the open interval \((a, b)\).
3. **Equal Endpoints**: \( f(a) = f(b) \).

Then, there exists a point \( c \in (a, b) \) such that \( f'(c) = 0 \).

### Step 1: Check Continuity and Differentiability
The function given is a logarithmic function. Since logarithmic functions are continuous and differentiable where the argument inside the logarithm is positive, we must ensure that:

\[
\frac{x^2 + ab}{x(a + b)} > 0
\]

For \( a > 0 \) and \( b > 0 \), the denominator \( x(a + b) \) is positive for \( x > 0 \), and the numerator \( x^2 + ab \) is positive for all \( x \). Hence, the argument of the logarithm is positive for \( x > 0 \), meaning that the function is continuous and differentiable on \( (a, b) \).

### Step 2: Check Equal Endpoints
We need to compute \( f(a) \) and \( f(b) \).

\[
f(a) = \log \left( \frac{a^2 + ab}{a(a + b)} \right)
= \log \left( \frac{a(a + b)}{a(a + b)} \right)
= \log(1) = 0
\]

Similarly,

\[
f(b) = \log \left( \frac{b^2 + ab}{b(a + b)} \right)
= \log \left( \frac{b(a + b)}{b(a + b)} \right)
= \log(1) = 0
\]

Thus, \( f(a) = f(b) = 0 \), satisfying the third condition of Rolle's theorem.

### Step 3: Find \( c \in (a, b) \) such that \( f'(c) = 0 \)
Now, we need to compute the derivative \( f'(x) \) and find \( c \) such that \( f'(c) = 0 \).

First, let:

\[
g(x) = \frac{x^2 + ab}{x(a + b)}
\]

Then,

\[
f(x) = \log(g(x))
\]

Using the chain rule:

\[
f'(x) = \frac{1}{g(x)} \cdot g'(x)
\]

We first compute \( g'(x) \). Differentiating \( g(x) \):

\[
g(x) = \frac{x^2 + ab}{x(a + b)}
\]

Apply the quotient rule:

\[
g'(x) = \frac{(2x)(x(a+b)) - (x^2 + ab)(a+b)}{[x(a+b)]^2}
\]

Simplifying the numerator:

\[
= \frac{2x^2(a+b) - (x^2 + ab)(a+b)}{x^2(a+b)^2}
= \frac{(a+b)(2x^2 - x^2 - ab)}{x^2(a+b)^2}
= \frac{(a+b)(x^2 - ab)}{x^2(a+b)^2}
\]

Thus,

\[
g'(x) = \frac{x^2 - ab}{x^2(a+b)}
\]

Now, we can substitute into the expression for \( f'(x) \):

\[
f'(x) = \frac{1}{g(x)} \cdot g'(x) = \frac{x(a+b)}{x^2 + ab} \cdot \frac{x^2 - ab}{x^2(a+b)} = \frac{x^2 - ab}{x^2 + ab}
\]

### Step 4: Solve \( f'(x) = 0 \)
Set \( f'(x) = 0 \):

\[
\frac{x^2 - ab}{x^2 + ab} = 0
\]

This implies:

\[
x^2 = ab
\]

Thus,

\[
x = \sqrt{ab}
\]

### Conclusion:
By Rolle's Theorem, there exists a point \( c = \sqrt{ab} \) in the interval \( (a, b) \) such that \( f'(c) = 0 \).

Would you like further details on any step? Here are some related questions:

1. How do you derive the quotient rule?
2. Why is logarithmic differentiation useful in this problem?
3. What would happen if \( f(a) \neq f(b) \)?
4. How can you interpret Rolle's Theorem geometrically?
5. What other conditions might invalidate Rolle's Theorem?

**Tip**: When applying Rolle's Theorem, always ensure that the function satisfies both the continuity and differentiability conditions.

Verify whether Rolle's theorem applies and find an appropriate constant c for the function f(x) = log((x^2 + ab) / (x(a+b))) on the interval [a, b], where a > 0.

Explore the application of Rolle's Theorem on the logarithmic function f(x) = log((x^2 + ab) / (x(a+b))) over the interval [a, b]. This detailed solution includes step-by-step differentiation, verification of conditions, and the determination of the constant c.

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