Solve in $\mathbb{R}$ the following equation: $4^x = \log_2(x) + \sqrt{x-1} + 14.$
My approach: I noticed that $x=2$ satisfies the equation, then I investigated the intervals $[1,2)$ and $(2,\infty)$, but it didn't lead me to a solution. Any help is appreciated.
Define $g(x)=4^x-\log_2(x)-\sqrt{x-1}-14$.
We have that $g'(x)=\log(4)4^x-\dfrac{1}{\log(2)x}-\dfrac{1}{2\sqrt{x-1}}$.
For $x\geq 2$, it follows that $$g'(x)\geq 4^2-\dfrac{1}{\log(2)}-\dfrac{1}{2}>0,$$
so $g$ is strictly increasing and cannot have any more roots on $(2,\infty)$.
As $\log_2(x)+\sqrt{x-1}+14\geq 14$ in $[1,2)$, there cannot be any roots before $\log_4(14)\approx 1.90$, in particular before $\dfrac{3}{2}$.
If there were a root on $\left(\dfrac{3}{2},2\right)$, by Rolle's Theorem there would exist some $c\in \left(\dfrac{3}{2},2\right)$ such that $g'(c)=0$.
But in $\left(\dfrac{3}{2},2\right)$ we have $$g'(x)\geq 4^{3/2}-\dfrac{1}{\log(2)}-\dfrac{1}{2\sqrt{0.5}}>0,$$
so there cannot be any roots on $\left(\dfrac{3}{2},2\right)$ and we conclude the only root of $g$ is $2$.