The function g is defined by \(\mathrm{g(x) = a \cdot b^x}\), where a and b are constants. If \(\mathrm{g(0) =...

1. TRANSLATE the given conditions into equations

• Given information:
  • \(\mathrm{g(x) = a \cdot b^x}\) (general exponential function form)
  • \(\mathrm{g(0) = 6}\) (function value when \(\mathrm{x = 0}\))
  • \(\mathrm{g(1) = 18}\) (function value when \(\mathrm{x = 1}\))

• What this tells us: We can substitute these x-values into our function to create equations for finding \(\mathrm{a}\) and \(\mathrm{b}\).

2. INFER the strategic approach

• Key insight: Since \(\mathrm{g(0)}\) involves \(\mathrm{b^0 = 1}\), this condition will directly give us the value of parameter \(\mathrm{a}\)
• Strategy: Solve for \(\mathrm{a}\) first using \(\mathrm{g(0) = 6}\), then use that result with \(\mathrm{g(1) = 18}\) to find \(\mathrm{b}\)

3. TRANSLATE the first condition

• Substitute \(\mathrm{x = 0}\) into \(\mathrm{g(x) = a \cdot b^x}\):
  \(\mathrm{g(0) = a \cdot b^0 = a \cdot 1 = a}\)

• Since \(\mathrm{g(0) = 6}\): \(\mathrm{a = 6}\)

4. TRANSLATE and SIMPLIFY the second condition

• Substitute \(\mathrm{x = 1}\) and \(\mathrm{a = 6}\) into \(\mathrm{g(x) = a \cdot b^x}\):
  \(\mathrm{g(1) = 6 \cdot b^1 = 6b}\)

• Since \(\mathrm{g(1) = 18}\): \(\mathrm{6b = 18}\)
• SIMPLIFY: \(\mathrm{b = 18 \div 6 = 3}\)

5. Verify the solution

• Check \(\mathrm{g(0)}\): \(\mathrm{6 \cdot 3^0 = 6 \cdot 1 = 6}\) ✓
• Check \(\mathrm{g(1)}\): \(\mathrm{6 \cdot 3^1 = 6 \cdot 3 = 18}\) ✓

Answer: 3

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students attempt to set up a system of two equations with two unknowns without recognizing that \(\mathrm{g(0) = 6}\) immediately gives them \(\mathrm{a = 6}\).

They might write: \(\mathrm{a \cdot b^0 = 6}\) and \(\mathrm{a \cdot b^1 = 18}\), then try to solve this as a complex system rather than recognizing that \(\mathrm{b^0 = 1}\) makes the first equation simply \(\mathrm{a = 6}\). This leads to unnecessary complexity and potential calculation errors, causing confusion and possibly guessing among the answer choices.

Second Most Common Error:

Missing conceptual knowledge about exponents: Students forget that \(\mathrm{b^0 = 1}\), so they treat \(\mathrm{g(0) = a \cdot b^0}\) as if it still contains the unknown \(\mathrm{b}\).

This prevents them from finding \(\mathrm{a = 6}\) in the first step, making it impossible to proceed systematically. This causes them to get stuck and abandon the systematic solution approach.

The Bottom Line:

This problem tests whether students can recognize the strategic advantage of using the condition with \(\mathrm{x = 0}\) first, since any base raised to the zero power equals 1, immediately simplifying one of the conditions.