Question

The number of non-negative values of n for which \(\log_{1/4}(n^2 - 7n + 14)>0\) is ____.

Hint:

Since the base $\tfrac14$ is less than 1, flip the inequality when removing the log: you need $0 \lt n^2-7n+14 \lt 1$. To check whether a quadratic can ever dip below a target value, shift it by that amount and look at the sign of its discriminant -- a negative discriminant with a positive leading coefficient means the expression never changes sign.

Answer 1

Correct Answer: 0

Approach: With a base between 0 and 1, the log is positive only when its argument is squeezed strictly between 0 and 1. Then just ask whether the quadratic ever dips below 1.

Step 1: We need \(\log_{1/4}(n^2 - 7n + 14) \gt 0\). Since \(0 \lt \tfrac14 \lt 1\), \(\log_{1/4}(A) \gt 0\) exactly when \(0 \lt A \lt 1\) (the base being less than 1 flips the usual sense). Here \(A = n^2 - 7n + 14\), so we need \[ 0 \lt n^2 - 7n + 14 \lt 1. \]

Step 2: Find the minimum of \(f(n) = n^2 - 7n + 14\). Vertex at \(n = \dfrac{7}{2} = 3.5\): \[ f(3.5) = 12.25 - 24.5 + 14 = 1.75. \]

Step 3: The parabola opens upward with minimum \(1.75\). So \(f(n) \ge 1.75 \gt 0\) for every \(n\) (left half of the requirement is always fine), but \(f(n)\) can never be below 1 (right half \(f(n) \lt 1\) is impossible).

Step 4: No \(n\) at all satisfies \(f(n) \lt 1\), so no non-negative \(n\) works either.

Final Answer: The count is \(\boxed{0}\).

Answer 2

Approach: Since the base \(\tfrac14\) is less than \(1\), \(\log_{1/4}(y)>0 \iff 0<y<1\). Find the minimum possible value of \(y=n^2-7n+14\) by completing the square, and check whether it can ever dip below \(1\).

Complete the square: \[ n^2-7n+14=\left(n-\frac72\right)^2+14-\frac{49}{4}=\left(n-\frac72\right)^2+\frac{7}{4}. \]
The minimum value of this expression (at \(n=3.5\)) is \(\tfrac74=1.75\), already greater than \(1\). Since \(y\geq1.75\) for every real \(n\), the condition \(y<1\) can never be satisfied, meaning \(\log_{1/4}(y)\) is never positive.

So there are no real values of \(n\) at all, let alone non-negative ones, that make the inequality true: \[ \boxed{0} \]

Answer 3

Concept:

• For a base $b$ with $0\lt b\lt 1$, $\log_b A \gt 0$ exactly when $0\lt A\lt 1$, because a fractional base reverses the usual direction of a logarithmic inequality.
• Instead of locating the vertex of a quadratic to find its minimum value, whether a quadratic can ever go below a target value can be checked directly from the discriminant of the shifted quadratic.

Step 1: Set up the required range for the argument.
Since $0 \lt \tfrac14 \lt 1$, $\log_{1/4}(A) \gt 0 \iff 0 \lt A \lt 1$, where $A = n^2-7n+14$. So we need $n^2-7n+14 \lt 1$ (the left part $A\gt0$ will be checked automatically once this is settled).

Step 2: Shift the quadratic and define $g(n) = n^2-7n+14-1 = n^2-7n+13$.
The inequality $n^2-7n+14\lt1$ is exactly the same as $g(n) \lt 0$.

Step 3: Test whether $g(n)$ can ever be negative, using its discriminant.
$D = (-7)^2 - 4(1)(13) = 49 - 52 = -3$
Since $D\lt0$ and the leading coefficient of $g$ is positive, $g(n) \gt 0$ for every real $n$ -- the parabola never touches or crosses the $n$-axis.

Step 4: Conclude for every real $n$.
Since $g(n)$ is never negative, the condition $n^2-7n+14 \lt 1$ is never true for any real value of $n$, so the original inequality has no solutions at all -- not among positive $n$, negative $n$, or non-negative $n$.

Final Answer: $0$