The function f is defined by \(\mathrm{f(x) = m \log_n(x) + k}\), where m and k are constants and n...

1. TRANSLATE the problem information

• Given information:
  • Function: \(\mathrm{f(x) = m \log_n(x) + k}\)
  • Point 1: \(\mathrm{(1, -4)}\) means \(\mathrm{f(1) = -4}\)
  • Point 2: \(\mathrm{(n^2, 10)}\) means \(\mathrm{f(n^2) = 10}\)
  • We need to find: \(\mathrm{m - k}\)

2. TRANSLATE the first point into an equation

• Substitute \(\mathrm{(1, -4)}\) into the function:
  \(\mathrm{-4 = m \log_n(1) + k}\)

3. INFER which logarithm property applies

• Since \(\mathrm{\log_n(1) = 0}\) for any valid base \(\mathrm{n \gt 1}\), the equation becomes:
  \(\mathrm{-4 = m(0) + k}\)
  \(\mathrm{-4 = k}\)

• Now we know \(\mathrm{k = -4}\)

4. TRANSLATE the second point into an equation

• Substitute \(\mathrm{(n^2, 10)}\) and \(\mathrm{k = -4}\) into the function:
  \(\mathrm{10 = m \log_n(n^2) + (-4)}\)
  \(\mathrm{14 = m \log_n(n^2)}\)

5. INFER which logarithm property to use next

• Since \(\mathrm{\log_n(n^2) = 2}\) (using the power property where \(\mathrm{\log_n(n^a) = a}\)):
  \(\mathrm{14 = m(2)}\)
  \(\mathrm{m = 7}\)

6. SIMPLIFY to find the final answer

• Calculate \(\mathrm{m - k}\):
  \(\mathrm{m - k = 7 - (-4) = 7 + 4 = 11}\)

Answer: C. 11

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that \(\mathrm{\log_n(n^2) = 2}\), instead trying to work with it as an unknown expression.

Without this key insight, they get stuck with the equation \(\mathrm{14 = m \log_n(n^2)}\) and can't solve for \(\mathrm{m}\). This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly find \(\mathrm{m = 7}\) and \(\mathrm{k = -4}\), but make an arithmetic error when calculating \(\mathrm{m - k = 7 - (-4)}\).

They might compute this as \(\mathrm{7 - 4 = 3}\) instead of \(\mathrm{7 + 4 = 11}\). This leads them to select Choice B (3).

The Bottom Line:

This problem tests whether students can apply logarithm properties strategically. The key breakthrough moments are recognizing that \(\mathrm{\log_n(1) = 0}\) and \(\mathrm{\log_n(n^2) = 2}\). Without these insights, students can't convert the points into solvable equations.