$\ln \frac{\beta}{\alpha}$

Let $f(x)=e^{x}$ . The given expression can be written as $k = \frac{f(b) - f(a)}{b-a}$ Note that $f(x)$ is continuous in the closed interval $[0,1]$ and differentiable in the open interval $(0,1)$ . By the Mean Value Theorem , there exists a number $c\in(a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$ It follows that there exists a number $c$ such that $f'(c) = k$ . Thus, $k=7e^{7x}$ . Given that $e^{7x}$ is monotonically increasing in the interval $(a,b)$ , and in fact over $\mathbb{R}$ , the minimum and maximum values of $k$ lie in the endpoints of the interval. Thus, $\begin{aligned} \alpha &= \min k = 7e^0 = 7 \\ \beta &= \max k = 7e^7 = 7e^7 \\ \end{aligned} \\ \boxed{\ln \left(\frac \beta \alpha \right) =7}$

Note: $e^{7x}$ is monotonically increasing since $\frac{d}{dx} \left(e^{7x}\right)>0$ for all $x\in\mathbb{R}$ .