\(4^{(k-3)} + 8 = 2\)How many distinct real solutions does the given equation have?

1. SIMPLIFY the equation by isolating the exponential term

• Given equation: \(4^{(k-3)} + 8 = 2\)
• Subtract 8 from both sides:
  \(4^{(k-3)} = 2 - 8\)
  \(4^{(k-3)} = -6\)

2. INFER whether this equation can have real solutions

• Key insight: Since the base is \(4 \gt 0\), the exponential function \(4^{(k-3)}\) is always positive for any real number k
• We need \(4^{(k-3)} = -6\), but this requires a positive quantity to equal a negative number
• This is mathematically impossible

3. APPLY CONSTRAINTS of real number solutions

• Since no positive number can equal any negative number, this equation has no solution in the real numbers

Answer: (D) Zero

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the fundamental property that exponential functions with positive bases are always positive. Instead, they attempt to solve \(4^{(k-3)} = -6\) by taking logarithms of both sides, not realizing that the logarithm of a negative number is undefined in real numbers. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Algebraic mistake during SIMPLIFY: Students make calculation errors when isolating the exponential term, perhaps getting \(4^{(k-3)} = 10\) instead of \(4^{(k-3)} = -6\). With a positive result, they might proceed to find a real solution using logarithms. This may lead them to select Choice (A) (Exactly one) thinking they found a valid solution.

The Bottom Line:

This problem tests whether students understand that exponential functions with positive bases have a restricted range (only positive outputs). The algebraic work is straightforward, but the conceptual insight about exponential function properties is what determines success.