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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 Forget it. Some minor confusion due to simultaneous editing. Now it is all fine!2012-06-26

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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!2012-06-26
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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 $f(b)-f(a)$.

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    @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.2012-06-27