Base and Exponent Variables

$\begin{aligned} x^{2x^6} & = 3 & \small \color{#3D99F6} \text{Let }x = 3^u \\ 3^{2u\cdot 3^{6u}} & = 3 \\ \implies 2u\cdot 3^{6u} & = 1 & \small \color{#3D99F6} \text{Taking log both sides} \\ \log 2 + \log u + 6u \log 3 & = 0 \end{aligned}$

For the $LHS = 0$ , $\implies \log u < 0$ or $u < 1$ . Let $u = \dfrac 1t$ , where $t > 1$ . Then we have:

$\begin{aligned} \log 2 - \log t + \frac 6t \log 3 & = 0 \\ \log 2 + \frac 6t \log 3 & = \log t & \small \color{#3D99F6} \text{Note that }t = 6 \text{ satisfies the equation.} \\ \log 2 + \log 3 & = \log 6 \end{aligned}$

Therefore, $x = 3^u = 3^\frac 1t = \sqrt[6] 3$ . $\implies a+b = 6+3 = \boxed 9$ .

Note that $x=-\sqrt[6]3$ is also a solution.