100 followers Problem

Consider the following

$\begin{aligned} 4\cos (e^x) & = 2^x + 2^{-x} = e^{x\ln 2} + e^{-x\ln 2} & \small \color{#3D99F6} \text{Divide both sides by }4 \\ \cos (e^x) & = \frac {e^{x\ln 2} + e^{-x\ln 2}}4 & \small \color{#3D99F6} \text{Let }u = e^x \\ \cos u & = \frac 14 \left(u^{\ln 2} + u^{-\ln 2}\right) & \small \color{#3D99F6} \text{where }u \in (0, \infty) \end{aligned}$

Since $e^x > 0$ for all $x$ , the domain of $u$ is $(0, \infty)$ . We note that the LHS $\cos u \in [-1, 1]$ . From AM-GM inequality, the RHS has a minimum of $\frac 12$ , since $u^{\ln 2} + u^{-\ln 2} \ge 2$ , and equality occurs when $u=1$ . Therefore, the RHS $\in [\frac 12, \infty)$ and it is $\le 1$ , where the solutions exist, when

$\begin{aligned} \frac 14 \left(u^{\ln 2} + u^{-\ln 2}\right) & \le 1 \\ u^{\ln 2} + u^{-\ln 2} & \le 4 \\ u^{2\ln 2} - 4 u^{\ln 2} + 1 \le 0 \\ \left(u^{\ln 2} - 2 - \sqrt 3\right) \left(u^{\ln 2} - 2 + \sqrt 3\right) & \le 0 \end{aligned}$

$\implies 0.1496 \le u \le 6.6859$ when the RHS is $\le 1$ . Since the range of $u$ , $6.6859-0.1496 = 6.5363 > 2\pi$ , there are $\boxed 4$ solutions as confirmed by the graph below (LHS = blue line, RHS = red line).