Key Concepts and Formulas

• Mean Value Theorem (MVT): If a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
• Derivative of $\log_e x$ (or $\ln x$): $\frac{d}{dx}(\log_e x) = \frac{1}{x}$.
• Change of Base Formula for Logarithms: $\log_a b = \frac{1}{\log_b a}$.

Step-by-Step Solution

Step 1: Verify the conditions for the Mean Value Theorem

We are given the function $f(x) = \log_e x$ and the interval $[1, 3]$. We need to check if $f(x)$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$.

• $f(x) = \log_e x$ is continuous for all $x > 0$. Since $[1, 3]$ contains only positive values, $f(x)$ is continuous on $[1, 3]$.
• The derivative of $f(x)$ is $f'(x) = \frac{1}{x}$, which exists for all $x \neq 0$. Since $(1, 3)$ does not contain $0$, $f(x)$ is differentiable on $(1, 3)$.

Since both conditions are satisfied, we can apply the Mean Value Theorem.

Step 2: Apply the Mean Value Theorem formula

The Mean Value Theorem states that there exists a $c \in (1, 3)$ such that $f'(c) = \frac{f(3) - f(1)}{3 - 1}$

Step 3: Calculate $f(3)$ and $f(1)$

We have $f(x) = \log_e x$. Therefore, $f(3) = \log_e 3$ $f(1) = \log_e 1 = 0$

Step 4: Calculate $f'(c)$

The derivative of $f(x) = \log_e x$ is $f'(x) = \frac{1}{x}$. Therefore, $f'(c) = \frac{1}{c}$

Step 5: Substitute into the MVT formula and solve for $c$

Substituting the calculated values into the MVT formula: $\frac{1}{c} = \frac{\log_e 3 - 0}{3 - 1}$ $\frac{1}{c} = \frac{\log_e 3}{2}$ $c = \frac{2}{\log_e 3}$

Step 6: Rewrite $c$ using the change of base formula

We want to express $c$ in terms of $\log_3 e$. Using the change of base formula, we know that $\log_e 3 = \frac{1}{\log_3 e}$. Substituting this into our expression for $c$: $c = \frac{2}{\frac{1}{\log_3 e}}$ $c = 2 \log_3 e$

Step 7: Verify that $c$ lies in the interval $(1, 3)$

We need to check if $1 < 2\log_3 e < 3$. We know $e \approx 2.718$. Since $3^0 = 1 < e < 3 = 3^1$, we have $0 < \log_3 e < 1$. More precisely, $\log_3 e \approx 0.91$. Thus, $c = 2 \times \log_3 e \approx 2 \times 0.91 = 1.82$. Since $1 < 1.82 < 3$, the value of $c$ we found is indeed within the open interval $(1, 3)$.

Step 8: Match with the given options

We found $c = 2 \log_3 e$, which corresponds to option (A).

Common Mistakes & Tips

• Forgetting to Check Conditions: Always verify that the function satisfies the continuity and differentiability conditions before applying the Mean Value Theorem.
• Incorrectly Calculating Derivatives: Be careful when calculating derivatives, especially of logarithmic functions.
• Not Simplifying the Answer: Make sure to simplify your answer and express it in a form that matches one of the given options.

Summary

We applied the Mean Value Theorem to the function $f(x) = \log_e x$ on the interval $[1, 3]$. By verifying the conditions, calculating the necessary function values and the derivative, and then solving for $c$, we found that $c = 2\log_3 e$. This corresponds to option (A).

The final answer is $\boxed{2\,{\log _3}e}$, which corresponds to option (A).