Find 'c' of Lagrange's Mean Value Theorem for the functions :
`f(x)=logx` in the interval [1, e]

Answer

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

### Step 1: Verify the conditions of LMVT
Lagrange's Mean Value Theorem states that if a function \( f \) 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 \( f(x) = \log x \):
- The function is continuous on \([1, e]\) and differentiable on \((1, e)\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Here, \( a = 1 \) and \( b = e \):
- \( f(1) = \log(1) = 0 \)
- \( f(e) = \log(e) = 1 \)

### Step 3: Calculate the difference quotient
Now, we can compute the difference quotient:

\[ \frac{f(b) - f(a)}{b - a} = \frac{f(e) - f(1)}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1} \]

### Step 4: Find the derivative \( f'(x) \)
Next, we need to find the derivative of \( f(x) \):

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

### Step 5: Set up the equation for \( c \)
According to LMVT, we set \( f'(c) \) equal to the difference quotient:

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

### Step 6: Solve for \( c \)
Now, we can solve for \( c \):

\[ \frac{1}{c} = \frac{1}{e - 1} \]

Taking the reciprocal of both sides gives:

\[ c = e - 1 \]

### Conclusion
Thus, the value of \( c \) that satisfies Lagrange's Mean Value Theorem for the function \( f(x) = \log x \) in the interval \([1, e]\) is:

\[ c = e - 1 \]