Find the value of c in Lagrange's mean value theorem for the function `f (x) = log _(e) x` on [1,2].

Options

A. `log 2`
B. `1-e`
C. `log _(2) e`
D. `1/e`

Answer

To find the value of \( c \) in Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = \log_e x \) on the interval \([1, 2]\), we will follow these steps:

### Step 1: Verify the conditions of LMVT
Lagrange's Mean Value Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

For our function \( f(x) = \log_e x \):
- **Continuity**: The function \( \log_e x \) is continuous for \( x > 0 \), hence it is continuous on \([1, 2]\).
- **Differentiability**: The function \( \log_e x \) is differentiable for \( x > 0 \), hence it is differentiable on \((1, 2)\).

Since both conditions are satisfied, we can apply LMVT.

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = 1 \) and \( b = 2 \):
- \( f(1) = \log_e(1) = 0 \)
- \( f(2) = \log_e(2) \)

### Step 3: Apply LMVT
Now, we can use the formula from LMVT:

\[ f'(c) = \frac{f(2) - f(1)}{2 - 1} \]

Substituting the values we found:

\[ f'(c) = \frac{\log_e(2) - 0}{2 - 1} = \log_e(2) \]

### Step 4: Find the derivative \( f'(x) \)
The derivative of \( f(x) = \log_e x \) is:

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

### Step 5: Set \( f'(c) \) equal to \( \log_e(2) \)
Now, we set \( f'(c) \) equal to \( \log_e(2) \):

\[ \frac{1}{c} = \log_e(2) \]

### Step 6: Solve for \( c \)
To find \( c \), we rearrange the equation:

\[ c = \frac{1}{\log_e(2)} \]

### Step 7: Rewrite using change of base formula
Using the change of base formula, we can express \( \log_e(2) \) as:

\[ \log_e(2) = \frac{1}{\log_2(e)} \]

Thus, we can write:

\[ c = \log_2(e) \]

### Final Answer
The value of \( c \) in Lagrange's Mean Value Theorem for the function \( f(x) = \log_e x \) on the interval \([1, 2]\) is:

\[ c = \frac{1}{\log_e(2)} \]