# 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.

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@Gigili There is no reason to capitalize 'V' and 'T' in 'Mean value theorem'. I rolled back your edit. –  user17762 Jun 26 '12 at 21:38
@Marvis: I edited the $\ln(x)$ part in the title and didn't get notified of your edit and saw it afterwards. So I rolled back to your version. Now I don't understand what you are saying. Feel free to do what you think is correct. –  Gigili Jun 26 '12 at 21:43
@Gigili Forget it. Some minor confusion due to simultaneous editing. Now it is all fine! –  user17762 Jun 26 '12 at 22:10

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 $\dfrac7{3 \ln(2)}$.

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Oh i see so i can just keep it as 3ln(2) on top i didnt know i can leave it like that. Thanks! –  soniccool Jun 26 '12 at 21:37
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 $f(b)-f(a)$.
@MartinSleziak: Thank you. I have loftier goals, am aiming for platinum in pre-calculus. Actually, teaching calculus can be kind of fun. Even classes of $200+$, which are more theatre than mathematics. –  André Nicolas Jun 27 '12 at 14:05