Question:

Step 1: Understand the given inequality and the conditions for the logarithmic function to be defined. The given inequality is $\log_{(x+1)/x}(a^2-3a+3)>0$, and for the logarithmic function to be defined, the base $(x+1)/x$ must be positive and not equal to 1, and the argument $a^2-3a+3$ must be positive. Step 2: Determine the conditions for the base of the logarithm to be valid. For $x > 0$, the base $(x+1)/x$ is always positive because both $x+1$ and $x$ are positive. To ensure the base is not equal to 1, we solve the equation $(x+1)/x = 1$. This simplifies to $x + 1 = x$, which has no solution for $x > 0$, meaning the base is always valid for $x > 0$. Step 3: Analyze the argument of the logarithm to ensure it is positive. The argument of the logarithm is $a^2 - 3a + 3$. For the logarithm to be defined, $a^2 - 3a + 3 > 0$. This is a quadratic inequality, and we need to find the values of $a$ for which this inequality holds true. Step 4: Solve the quadratic inequality $a^2 - 3a + 3 > 0$. To solve $a^2 - 3a + 3 > 0$, we look for the roots of the equation $a^2 - 3a + 3 = 0$. Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-3$, and $c=3$, we get $a = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(3)}}{2(1)} = \frac{3 \pm \sqrt{9 - 12}}{2} = \frac{3 \pm \sqrt{-3}}{2}$. Since the discriminant is negative, there are no real roots, and the quadratic expression $a^2 - 3a + 3$ is always positive for all real $a$. Step 5: Determine the condition for the logarithmic inequality to be greater than 0. For $\log_{(x+1)/x}(a^2-3a+3)>0$, given that $a^2 - 3a + 3 > 0$ is always true, we need $(x+1)/x > 1$ because $\log_b(a) > 0$ if and only if $a > 1$ and $b > 1$, or $0 < a < 1$ and $0 < b < 1$. Since $(x+1)/x$ is always greater than 1 for $x > 0$, we focus on ensuring $a^2 - 3a + 3 > 1$ for the logarithm to be positive. Step 6: Solve for $a$ in $a^2 - 3a + 3 > 1$. The inequality simplifies to $a^2 - 3a + 2 > 0$. Factoring the quadratic gives $(a-1)(a-2) > 0$. This inequality holds true when either both factors are positive or both are negative. Both factors are positive when $a > 2$, and both are negative when $a < 1$. Since we seek the least positive integral value of $a$, we consider $a > 2$. Step 7: Find the least positive integral value of $a$ satisfying the conditions. Given that $a$ must be an integer greater than 2, the least positive integral value of $a$ that satisfies all conditions is $a = 3$. Step 8: Conclude the final answer based on the given options. The final answer is: $\boxed{3}$