If the function `f(x)=logx, x in [1, e]`, satisfies all the conditions of LMVT, then the value of c is

Options

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

Answer

To solve the problem, we will apply the Mean Value Theorem (MVT) to the function \( f(x) = \log x \) over the interval \([1, e]\).

### Step-by-Step Solution:

1. **Identify the function and interval**:
The function given is \( f(x) = \log x \) and the interval is \([1, e]\).

2. **Check the conditions of MVT**:
- **Continuity**: The function \( f(x) = \log x \) is continuous on the interval \([1, e]\) since the logarithm is defined for all \( x > 0 \).
- **Differentiability**: The function \( f(x) = \log x \) is differentiable on the open interval \( (1, e) \).

3. **Apply the Mean Value Theorem**:
According to MVT, there exists at least one \( c \) in \( (1, e) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
where \( a = 1 \) and \( b = e \).

4. **Calculate \( f(a) \) and \( f(b) \)**:
- \( f(1) = \log 1 = 0 \)
- \( f(e) = \log e = 1 \)

5. **Calculate \( f(b) - f(a) \)**:
\[ f(e) - f(1) = 1 - 0 = 1 \]

6. **Calculate \( b - a \)**:
\[ b - a = e - 1 \]

7. **Substitute into the MVT formula**:
\[ f'(c) = \frac{1}{e - 1} \]

8. **Find the derivative \( f'(x) \)**:
The derivative of \( f(x) = \log x \) is:
\[ f'(x) = \frac{1}{x} \]

9. **Set \( f'(c) \) equal to the average rate of change**:
\[ \frac{1}{c} = \frac{1}{e - 1} \]

10. **Solve for \( c \)**:
Cross-multiplying gives:
\[ 1 \cdot (e - 1) = 1 \cdot c \implies c = e - 1 \]

### Final Answer:
The value of \( c \) is \( e - 1 \).

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