Correct Answer: A) $12$ Let the given equation be: $$\log_3 x + \log_x 9 = 3$$ We know the change of base formula for logarithms: $\log_b a = \frac{\log_c a}{\log_c b}$. Using this, we can express $\log_x 9$ in terms of base 3: $$\log_x 9 = \frac{\log_3 9}{\log_3 x}$$ Since $\log_3 9 = \log_3 (3^2) = 2$, we have: $$\log_x 9 = \frac{2}{\log_3 x}$$ Substitute this back into the original equation: $$\log_3 x + \frac{2}{\log_3 x} = 3$$ To simplify, let $y = \log_3 x$. The equation becomes: $$y + \frac{2}{y} = 3$$ Multiply the entire equation by $y$ (note that $y \neq 0$ because if $y=0$, $\log_3 x = 0 \implies x=1$, but $\log_x 9$ is undefined for $x=1$): $$y^2 + 2 = 3y$$ Rearrange it into a standard quadratic equation: $$y^2 - 3y + 2 = 0$$ Factor the quadratic equation: $$(y-1)(y-2) = 0$$ This gives two possible values for $y$: $$y = 1 \quad \text{or} \quad y = 2$$ Now, substitute back $y = \log_3 x$ to find the values of $x$: **Case 1: $y = 1$** $$\log_3 x = 1$$ $$x = 3^1 = 3$$ **Case 2: $y = 2$** $$\log_3 x = 2$$ $$x = 3^2 = 9$$ Both values $x=3$ and $x=9$ are valid because they satisfy the domain conditions for the original logarithmic expressions ($x > 0$ and $x \neq 1$). The question asks for the sum of all possible values of $x$: $$\text{Sum} = 3 + 9 = 12$$ Thus, the sum of all possible values of $x$ is $12$.