Solve the exponential sum (variant): If 2^(x−1) + 2^(x+1) = 320, determine x.

Introduction / Context:
This is the same structure as a previous item with a different right-hand side. We again factor a shared power of 2 to convert the equation to a simple power-of-two equality.

Given Data / Assumptions:

• 2^(x−1) + 2^(x+1) = 320
• Standard index laws apply.

Concept / Approach:
Factor 2^(x−1) and use 2^(x+1) = 2^(x−1)*2^2. Solve the resulting equation by recognizing powers of 2.

Step-by-Step Solution:
2^(x−1)(1 + 2^2) = 2^(x−1)*5 = 3202^(x−1) = 320/5 = 64 = 2^6x − 1 = 6 ⇒ x = 7

Verification / Alternative check:
Check: 2^6 + 2^8 = 64 + 256 = 320, correct.

Why Other Options Are Wrong:
Any value other than 7 fails to satisfy the exact equality.

Common Pitfalls:
Arithmetic slip dividing 320 by 5. Ensure 320/5 = 64.

Final Answer:
7