I recently encountered this problem on a problem set:
Given that $$\log_2(2 + \sqrt{3}) < \log(3\pi)$$ find the number of roots of the equation $$4\cos(e^x) = 2^x + 2^{-x}$$
I consulted my teacher for the solution and he checked the solution he had and told me that it was not possible to solve this manually as the solution he had was via graphing the functions and it was not possible without a graphing calculator or an app.
While using a graphing calculator the initial inequality is not used but it must have been given for a reason.
Graphically 4 solutions are obtained.
I want to know if there is really no way to solve this question using only calculations which are doable by hand.
Call the LHS and RHS of the equation
$L(x) = 4\cos(e^x)$
and
$R(x) = 2^x+2^{-x}$
respectively.
On the interval $[0,\ln{\pi}]$, $L(x)$ strictly decreases from $4\cos{1}\approx2.16$ to $-4$. On the same interval, $R(x)$ strictly increases from $2$ to $2^{\ln{\pi}}+2^{-\ln{\pi}}\approx2.66$. We then deduce that there is exactly one root in $[0,\ln{\pi}]$.
On the interval $[\ln{\pi},\ln{\frac{3}{2}\pi}]$, $L(x)\le0$ while $R(x)\gt0$ so there are zero roots in this interval.
On the interval $[\ln{\frac{3}{2}\pi},\ln{\frac{5}{2}\pi}]$, $L(x)$ is concave down ($L''\lt0$). On the same interval, $R(x)$ is concave up ($R''\gt0$). So there are either 0 or 2 roots in this interval. At $x=\ln{2\pi}$, $L(x)=4$ is greater than $R(x)=2^{\ln{2\pi}}+2^{-2\ln{\pi}}\approx3.85$. Yet on the ends of the interval $L(x) \lt R(x)$. We deduce that there are exactly 2 roots in this interval.
On the interval $[\ln{\frac{5}{2}\pi},\infty)$ there are no more roots because $R(x)>4$ and $L(x)\le4$.
For $x\lt0$ as $x\rightarrow-\infty$, $L(x)$ asymptotically approaches $4$ with continuously decreasing (negative) slope. In the same interval $R(x)$ is increasing with continuously increasing (negative) slope. We deduce that there is exactly one root in this interval.
For a total of four roots.