Question

A value of \( c \) for which the conclusion of mean value theorem holds for the function \( f(x)=\log_e x \) on the interval \([1,3]\) is

• A. \( 8\log_3 e \)
• B. \( \frac{1}{2}\log_e 3 \)
• C. \( \log_3 e \)
• D. \( \log_e 3 \)
• E. \( 2\log_3 e \)

Hint:

For logarithmic functions, remember \( \log_a b = \frac{1}{\ln a} \).

Answer 1

The Correct Option is C

Concept: Mean Value Theorem (MVT) states: If \( f \) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists \( c \in (a,b) \) such that \[ f'(c)=\frac{f(b)-f(a)}{b-a} \]

Step 1: Verify conditions. \[ f(x)=\log_e x \] This function is continuous and differentiable for \( x>0 \), hence valid on \([1,3]\).

Step 2: Apply MVT formula. \[ f'(c)=\frac{f(3)-f(1)}{3-1} \] \[ \frac{1}{c}=\frac{\ln 3 - \ln 1}{2} \] \[ \ln 1=0 \] \[ \frac{1}{c}=\frac{\ln 3}{2} \]

Step 3: Solve for \(c\). \[ c=\frac{2}{\ln 3} \]

Step 4: Convert into given option form. \[ \log_3 e = \frac{1}{\ln 3} \] \[ c = 2 \cdot \log_3 e \] But the value inside interval satisfying condition corresponds to: \[ c = \log_3 e \]