Jun 15, 2018

$b = 1 - \ln \frac{\ln 2}{\ln} 2$

#### Explanation:

The function is of the form

$f \left(x\right) = a x - {2}^{x}$

and so

${f}^{'} \left(x\right) = a - \left(\ln 2\right) {2}^{x}$

Thus ${f}^{'} \left(b\right) = 0 \implies$

$0 = a - \left(\ln 2\right) {2}^{b}$

${2}^{b} = \frac{a}{\ln} 2 \implies b = {\log}_{2} \left(\frac{a}{\ln} 2\right) = \ln \frac{\frac{a}{\ln} 2}{\ln} 2$

To complete the answer we need to determine the value of $a$ explicitly.

$a = \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + \ldots}}} \implies$

${a}^{3} = 6 + \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + \ldots}}} = 6 + a \implies$

${a}^{3} - a - 6 = 0 \implies$

${a}^{3} - 2 {a}^{2} + 2 {a}^{2} - 4 a + 3 a - 6 = 0 \implies$

$\left({a}^{2} + 2 a + 3\right) \left(a - 2\right) = 0$

Since ${a}^{2} + 2 a + 3 = 0$ can not have real roots, we must have $a = 2$

Thus

$b = \ln \frac{\frac{2}{\ln} 2}{\ln} 2 = \frac{\ln 2 - \ln \left(\ln 2\right)}{\ln 2} \implies$

$b = 1 - \ln \frac{\ln 2}{\ln} 2$