The function g is defined by \(\mathrm{g(x) = a\, ln(x) + b}\), where a and b are constants. The graph...

1. TRANSLATE the problem information

• Given information:
  • Function: \(\mathrm{g(x) = a \ln(x) + b}\)
  • Point 1: \(\mathrm{(e^2, 8)}\) means \(\mathrm{g(e^2) = 8}\)
  • Point 2: \(\mathrm{(e^3, 23)}\) means \(\mathrm{g(e^3) = 23}\)

• What this tells us: We can substitute these points into our function to create equations

2. SIMPLIFY using logarithm properties

• For point \(\mathrm{(e^2, 8)}\): \(\mathrm{g(e^2) = a \ln(e^2) + b = 8}\)
  Since \(\mathrm{\ln(e^2) = 2}\), we get: \(\mathrm{2a + b = 8}\)

• For point \(\mathrm{(e^3, 23)}\): \(\mathrm{g(e^3) = a \ln(e^3) + b = 23}\)
  Since \(\mathrm{\ln(e^3) = 3}\), we get: \(\mathrm{3a + b = 23}\)

3. INFER the solving strategy

• We now have a system of two equations with two unknowns:
  • \(\mathrm{2a + b = 8}\) ... (1)
  • \(\mathrm{3a + b = 23}\) ... (2)

• Since both equations have +b, elimination will be efficient

4. SIMPLIFY the system using elimination

• Subtract equation (1) from equation (2):
  \(\mathrm{(3a + b) - (2a + b) = 23 - 8}\)
  \(\mathrm{a = 15}\)

• Substitute \(\mathrm{a = 15}\) back into equation (1):
  \(\mathrm{2(15) + b = 8}\)
  \(\mathrm{30 + b = 8}\)
  \(\mathrm{b = -22}\)

Answer: -22

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Not correctly evaluating \(\mathrm{\ln(e^2)}\) and \(\mathrm{\ln(e^3)}\)

Students might treat \(\mathrm{\ln(e^2)}\) as just "\(\mathrm{\ln(e^2)}\)" without simplifying to 2, or incorrectly calculate \(\mathrm{\ln(e^2) = 2\ln(e) = 2(1) = 2}\). While this accidentally gives the right answer, some students might calculate \(\mathrm{\ln(e^2) = (\ln(e))^2 = 1^2 = 1}\), or make other logarithm property errors.

This leads to incorrect equations like \(\mathrm{a + b = 8}\) instead of \(\mathrm{2a + b = 8}\), resulting in wrong values for both a and b.

Second Most Common Error:

Poor INFER reasoning: Setting up equations correctly but choosing substitution over elimination

Students solve \(\mathrm{2a + b = 8}\) for \(\mathrm{b = 8 - 2a}\), then substitute into the second equation: \(\mathrm{3a + (8 - 2a) = 23}\), getting \(\mathrm{a + 8 = 23}\), so \(\mathrm{a = 15}\). While this works, it's more prone to algebra errors than elimination, and students often make sign errors during the substitution step.

The Bottom Line:

This problem tests whether students can bridge logarithm properties with linear systems. The key insight is recognizing that \(\mathrm{\ln(e^n) = n}\) immediately, then treating it as a straightforward system of equations problem.