Solve the equation: $\;$ $\log_3 \left[\sqrt{130 - 7^{\log_x \left(6 - x\right)}}\right] = 2$
Given equation: $\;\;$ $\log_3 \left[\sqrt{130 - 7^{\log_x \left(6 - x\right)}}\right] = 2$
i.e. $\;$ $\dfrac{1}{2} \log_3 \left[130 - 7^{\log_x \left(6 - x\right)}\right] = 2$
i.e. $\;$ $\log_3 \left[130 - 7^{\log_x \left(6 - x\right)}\right] = 4$
i.e. $\;$ $130 - 7^{\log_x \left(6 - x\right)} = 3^4 = 81$
i.e. $\;$ $7^{\log_x \left(6 - x\right)} = 49 = 7^2$
i.e. $\;$ $\log_x \left(6 - x\right) = 2$
i.e. $\;$ $6 - x = x^2$
i.e. $\;$ $x^2 + x - 6 = 0$
i.e. $\;$ $\left(x + 3\right) \left(x - 2\right) = 0$
i.e. $\;$ $x = -3$ $\;$ or $\;$ $x = 2$
But $\;$ $x = -3$ $\;$ makes the base of the term $\;$ $\log_x \left(6 - x\right)$, $\;$ in the given question, negative.
$\therefore \;$ $x = -3$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$