So $ 2^{n-1} > 166.67 $ implies $ n-1 \geq 8 $, thus $ n \geq 9 $.

["Understanding Why $ 2^{n-1} > 166.67 $ Implies $ n \geq 9 $: A Step-by-Step Explanation", "When solving inequalities involving exponents, clear reasoning is key to avoiding mistakes. One common inequality that students encounter is:", "> $ 2^{n-1} > 166.67 $ implies $ n - 1 \geq 8 $, thus $ n \geq 9 $", "At first glance, this step might seem straightforward, but breaking it down reveals important mathematical principles. This article explains why this implication holds true and helps readers understand how to correctly interpret exponential inequalities.", "---", "### What Does $ 2^{n-1} > 166.67 $ Mean?", "The expression $ 2^{n-1} $ grows exponentially as $ n - 1 $ increases. To solve $ 2^{n-1} > 166.67 $, we need to find the smallest integer $ n $ such that this inequality holds.", "First, note that $ 166.67 $ is approximately $ \frac{500}{3} $ or closely related to $ \frac{2^7}{1.25} $, but working with the decimal keeps the reasoning intuitive:", "$$\n2^{n-1} > 166.67\n$$", "We compare powers of 2:", "- $ 2^7 = 128 $\n- $ 2^8 = 256 $", "Since $ 128 < 166.67 < 256 = 2^8 $, we deduce:", "$$\n2^8 > 166.67 \quad \Rightarrow \quad n - 1 \geq 8\n$$", "---", "### Why Does $ n - 1 \geq 8 $ Based on $ 2^{n-1} > 166.67 $?", "From $ 2^{n-1} > 166.67 $ and knowing $ 2^8 = 256 $, we infer that $ n - 1 $ must be at least 8 to satisfy the inequality, since:", "- If $ n - 1 < 8 $, then $ 2^{n-1} < 256 $ but could still be less than or equal to 166.67 (e.g., $ n - 1 = 7 $ gives $ 128 < 166.67 $).\n- Only when $ n - 1 \geq 8 $ does $ 2^{n-1} \geq 256 > 166.67 $.", "To be precise:", "$$\nn - 1 \geq 8 \quad \Rightarrow \quad n \geq 9\n$$", "This inclusion is critical: $ n - 1 \geq 8 $ means $ n $ is at least 9, not strictly greater than 9, so the smallest valid integer value for $ n $ is exactly $ n = 9 $.", "---", "### Common Pitfall: Assuming Strict Inequality Without Checking Powers of 2", "A frequent error is skipping the exact powers:\nSince $ 2^7 = 128 < 166.67 $ and $ 2^8 = 256 > 166.67 $, simply stating $ n - 1 > \log_2(166.67) $ might mislead if not tested against integer exponents. Only testing $ 2^8 $ confirms $ n - 1 \geq 8 $ is needed.", "---", "### How to Solve It Correctly: Step-by-Step", "1. Start with:\n $$\n 2^{n-1} > 166.67\n $$\n2. Estimate the smallest power of 2 exceeding 166.67:\n $ 2^8 = 256 $, $ 2^7 = 128 $ → so $ n - 1 \geq 8 $\n3. Solve:\n $$\n n - 1 \geq 8\n $$\n4. Add 1 to both sides:\n $$\n n \geq 9\n $$", "---", "### Real-World Implication in Problem-Solving", "Understanding this inequality helps in algorithmic complexity, growth modeling, and binary scaling—contexts where exponential functions frequently appear. Knowing $ 2^{n-1} > 166.67 $ implies $ n \geq 9 $ lets you deterministically determine the minimum input size required for a system exceeding a threshold.", "---", "### Summary", "- $ 2^{n-1} > 166.67 $ means the exponent $ n - 1 $ must satisfy $ 2^{n-1} > 166.67 $\n- $ 2^8 = 256 > 166.67 $, so $ n - 1 \geq 8 $ is necessary\n- Adding 1 gives $ n \geq 9 $\n- This step ensures correct integer reasoning without assuming tighter bounds", "Mastering such inequalities strengthens logical thinking and is essential in math and computer science disciplines.", "---", "### Key Takeaway", "> Because $ 2^8 = 256 $ is the smallest power of 2 strictly greater than 166.67, and $ 2^7 = 128 $ falls short, $ n - 1 \geq 8 $ follows, leading directly to $ n \geq 9 $. This precise step-by-step validation ensures accuracy when solving exponential inequalities.", "---", "Keywords: $ 2^{n-1} > 166.67 $, $ n - 1 \geq 8 $, $ n \geq 9 $, exponential inequality, logarithmic reasoning, step-by-step math explanation, algebraic inequality, computing exponents, binary growth."]