⚡ AZUCATION SHORTCUT:
Base $< 1 \implies$ flip inequality: $0 < n^2 - 7n + 11 < 1$.
Check integers around $n = 3.5$ (vertex):
$n=3 \implies 9-21+11 = -1$ (False)
$n=2 \implies 4-14+11 = 1$ (False)
$n=4 \implies 16-28+11 = -1$ (False)
Wait, let's look closer at the range: $n$ must be 2 or 5? No, check $n=2$: $11-10=1$. Check $n=3$: $20-21=-1$. Only $n=2$ and $n=5$ give values between 0 and 1.

Step 1: Understand Logarithmic Inequalities
For $\log_a(x) > 0$, we have two cases based on the base $a$:
1. If $a > 1$, then $x > 1$.
2. If $0 < a < 1$, then $0 < x < 1$.
Here, the base is $\frac{1}{4}$, which is less than 1. So, we must solve $0 < n^2 - 7n + 11 < 1$.

Step 2: Solve the Right Inequality
$n^2 - 7n + 11 < 1 \implies n^2 - 7n + 10 < 0$.
Factoring the quadratic: $(n-2)(n-5) < 0$.
This holds true for values of $n$ in the range $(2, 5)$.

Step 3: Solve the Left Inequality (Domain Check)
$n^2 - 7n + 11 > 0$.
The roots of $n^2 - 7n + 11 = 0$ are $\frac{7 \pm \sqrt{49 - 44}}{2} = \frac{7 \pm \sqrt{5}}{2}$.
$\sqrt{5} \approx 2.23$, so roots are $\approx 2.38$ and $\approx 4.62$.
The expression is positive when $n < 2.38$ or $n > 4.62$.

Step 4: Find Integer Intersections
From Step 2, $n$ can be integers 3 or 4.
From Step 3, $n$ must be $< 2.38$ or $> 4.62$.
Testing $n=3$: $3^2 - 7(3) + 11 = 9 - 21 + 11 = -1$ (Does not satisfy $x > 0$).
Testing $n=4$: $4^2 - 7(4) + 11 = 16 - 28 + 11 = -1$ (Does not satisfy $x > 0$).
Actually, the integer solutions must be within $(2, 5)$ but outside $(2.38, 4.62)$. There are 0 such integers.

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