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.
 A: Let $f(x)=\ln x$. Then there is a $c$ between $1$ and $8$ such that 
$$\frac{f(8)-f(1)}{8-1}=f'(c)=\frac{1}{c}.$$
So we have $\dfrac{\ln 8}{7}=\dfrac{1}{c}$. (Here we used the fact that $\ln 1=0$.)
Flip both fractions over. We get $c=\dfrac{7}{\ln 8}$.
Remark: In most cases, one cannot solve explicitly for $c$. That's not the point of the Mean Value Theorem.  What is useful about MVT is that if you know something about the size of the derivative, it tells you something about the size of the change $f(b)-f(a)$. 
A: 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 $\dfrac7{3 \ln(2)}$.
