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 $1/3 \le z < w \le 3 \Big)$ $$\displaystyle{ | \ln w - \ln z | \over | w-z | } \le 3 \ \ \ \ \longrightarrow$$ $$| \ln w - \ln z | \le 3 | w-z |$$