calculus - Mean value theorem with $ln(x)$

I understand how to do mean value theorum but I'm not sure how to apply it with $\ln(x)$.

$$f(x) = \ln(x), \ [1, 8]$$

How can I find a $c$ that satisfies the conclusion of the Mean Value theorem by using $\ln(x)$?

I know its $\dfrac{f(b)-f(a)}{b-a}$, then take derivative and fill in the slope.

But how do I solve this with ln? I only did this with quadratic.

Answer

The mean value theorem states that if $f(x)$ is continuous on an interval $[a,b]$ and differentiable on $(a,b)$, then there exists a $c \in (a,b)$ such that

$$f'(c) = \dfrac{f(b)-f(a)}{b-a}$$
In your case, the function $f(x) = \ln(x)$ is continuous on an interval $[1,8]$ and differentiable on $(1,8)$. The derivative of $\ln(x)$ is $\dfrac1{x}$ in the interval $(1,8)$. Hence, by mean value theorem, $\exists c \in (1,8)$ such that $$f'(c) = \dfrac1c = \dfrac{f(8)-f(1)}{8-1} = \dfrac{\ln(8) - \ln(1)}{8-1} = \dfrac{3 \ln(2) - 0}{7} = \dfrac{3 \ln(2)}7$$
Hence, the desired point $c$ is $\dfrac7{3 \ln(2)}$.