Exponential Piecewise equation

Given that:

$\begin{aligned} 8^{\left|2\lceil x \rceil - 3\lfloor x \rfloor \right|} & = 4^{2\lfloor x \rfloor^2 - 5\lceil x \rceil + 3.5} \\ 2^{\left|6\lceil x \rceil - 9\lfloor x \rfloor \right|} & = 2^{4\lfloor x \rfloor^2 - 10\lceil x \rceil + 7} \\ \implies \left|6\lceil x \rceil - 9\lfloor x \rfloor \right| & = 4\lfloor x \rfloor^2 - 10\lceil x \rceil + 7 \end{aligned}$

If $x$ is an integer, then $\lceil x \rceil = \lfloor x \rfloor = x$ and $|-3x| = 4x^2 - 10x + 7$ or

$4x^2 - 10x + 7 = \begin{cases} 3x & \text{if }x \ge 0 & \implies 4x^2 - 13x + 7 = 0 & \text{No integer solution.} \\ - 3x & \text{if }x < 0 & \implies 4x^2 - 7x + 7 = 0 & \text{No real solution.} \end{cases}$

Therefore, there is no integer solution or $x$ is not an integer. Then $\lceil x \rceil = \lfloor x \rfloor + 1$ and $4\lfloor x \rfloor^2 - 10\lfloor x \rfloor - 3 = |6-3\lfloor x \rfloor|$ or

$\begin{cases} 4\lfloor x \rfloor^2 - 7\lfloor x \rfloor - 9 = 0 & \text{if }\lfloor x \rfloor \le 2 & \implies \text{No integer solution.} \\ 4\lfloor x \rfloor^2 - 13\lfloor x \rfloor + 3 = 0 & \text{if }\lfloor x \rfloor > 2 & \implies \lfloor x \rfloor = 3 \text{ as } \lfloor x \rfloor \ne \frac 14 \end{cases}$

Therefore, $3 < x < 4$ and $a+b = 3+4 = \boxed 7$ .

If $a=\lfloor{x}\rfloor$ and $b=\lceil{x}\rceil$ , we know that $b=a+1$ . Substituting this, we get $3|2(a+1)-3a|=2(2a^2-5(a+1)+3.5)$ . Solving the equation yields $4$ real solutions for $a$ , but only integral solutions work so $a=3$ . This means that $\lfloor{x}\rfloor=3$ and $\lceil{x}\rceil=4$ so $3<x<4$ . As a result, the final answer is $3+4=7$ .