SOLUTION 14: Consider the function $ f(x) = \ln x$ on the interval $ [1/3, 3] $, and assume that $ 1/3 \le z < w \le 3 $. This function is continuous (well-known) on the closed interval $ [z, w]$. The derivative of $f$ is
$$ f'(x) = \displaystyle{ 1 \over x } $$
We can now see that $f$ is differentiable on the open interval $ (z, w) $. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem on the closed interval $ [z, w] $. Then there is some number $c$, $ z < c < w$, so that
$$ f'(c) = \displaystyle{ f(w)-f(z) \over w-z } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over c } = \displaystyle{ \ln w - \ln z \over w-z } \ \ \ \ \longrightarrow $$
$$ \Bigg| \displaystyle{ \ln w - \ln z \over w-z } \Bigg| = \Big| \displaystyle{ 1 \over c } \Big| \le
\Big| \displaystyle{ 1 \over 1/3 } \Big| =3
\ \ \ \ \longrightarrow $$
$\Big($ since we assumed that $ 1/3 \le z < w \le 3 \Big)$
$$ \displaystyle{ | \ln w - \ln z | \over | w-z | } \le 3 \ \ \ \ \longrightarrow $$
$$ | \ln w - \ln z | \le 3 | w-z | $$
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