Solve the equation: $\;$ $\left(3^{x^2 - 7.2x + 3.9} - 9 \sqrt{3}\right) \log \left(7 - x\right) = 0$
Given equation: $\;\;$ $\left(3^{x^2 - 7.2x + 3.9} - 9 \sqrt{3}\right) \log \left(7 - x\right) = 0$ $\;\;\; \cdots \; (1)$
$\implies$ $3^{x^2 - 7.2x + 3.9} - 9 \sqrt{3} = 0$ $\;$ or $\;$ $\log \left(7 - x\right) = 0$
i.e. $\;$ $3^{x^2 - 7.2x + 3.9} = 9 \sqrt{3}$ $\;\;\; \cdots \; (2)$ $\;$ or $\;$ $\log \left(7 - x\right) = \log \left(1\right)$ $\;\;\; \cdots \; (3)$
Consider equation $(2)$.
We have, $\;$ $3^{x^2 - 7.2x + 3.9} = 3^2 \times 3^{\frac{1}{2}} = 3^{\frac{5}{2}}$
$\implies$ $x^2 - 7.2x + 3.9 = \dfrac{5}{2}$
i.e. $\;$ $x^2 - 7.2x + 1.4 = 0$
i.e. $\;$ $5x^2 - 36x + 7 = 0$
i.e. $\;$ $\left(x - 7\right) \left(5x - 1\right) = 0$
i.e. $\;$ $x = 7$ $\;$ or $\;$ $x = \dfrac{1}{5}$
When $\;$ $x = 7$, $\;$ equation $(1)$ will have the term $\;$ $\log \left(7 - 7\right) = \log 0$
But $\;$ $\log 0$ $\;$ is not defined.
Therefore, $x = 7$ is not a valid solution.
Consider equation $(3)$.
We have, $\;$ $7 - x = 1$ $\implies$ $x = 6$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{5}, \; 6\right\}$