To solve this problem using the Mean Value Theorem (MVT), we need to verify the hypotheses and then find $c$ that satisfies the conclusion.

Mean Value Theorem Statement:

If a function $f(x)$:

1. is continuous on a closed interval $[a, b]$,
2. and differentiable on the open interval $(a, b)$,

then there exists a number $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}.$

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Step 1: Verify Hypotheses

1. $f(x) = \ln(x)$ is continuous on $[1, 8]$ since $\ln(x)$ is continuous for $x > 0$.
2. $f(x)$ is differentiable on $(1, 8)$ since the derivative of $\ln(x)$ exists and is defined for $x > 0$.

Thus, $f(x)$ satisfies the hypotheses of the MVT.

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Step 2: Find $f'(c)$ Using MVT Conclusion

$f'(c) = \frac{f(b) - f(a)}{b - a}.$ Substitute $a = 1$, $b = 8$, and $f(x) = \ln(x)$: $\frac{f(8) - f(1)}{8 - 1} = \frac{\ln(8) - \ln(1)}{7}.$ Since $\ln(1) = 0$: $\frac{\ln(8)}{7}.$

Now, $f'(c)$ must equal this value: $f'(c) = \frac{\ln(8)}{7}.$

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Step 3: Solve for $c$

The derivative of $f(x) = \ln(x)$ is: $f'(x) = \frac{1}{x}.$ So: $\frac{1}{c} = \frac{\ln(8)}{7}.$ Solve for $c$: $c = \frac{7}{\ln(8)}.$

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Final Answer:

The value of $c$ that satisfies the conclusion of the Mean Value Theorem is: $c = \frac{7}{\ln(8)} \quad \text{where } c \in (1, 8).$

Would you like me to clarify or expand on any step?

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Related Questions:

1. What are the geometrical implications of the Mean Value Theorem?
2. How would the solution change if $f(x)$ were a different logarithmic function, e.g., $\ln(x + 1)$?
3. Can the Mean Value Theorem be applied if $f(x)$ is discontinuous at one point in the interval?
4. What happens if $c$ is outside the interval $(1, 8)$?
5. What if the interval were changed to $[1, e^2]$? What would $c$ be?

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Tip:

Always check the continuity and differentiability conditions before applying the Mean Value Theorem!