We need to solve: 0.05n log₂(n) < 0.01n².

["Title: Solving the Inequality: 0.05 n log₂(n) < 0.01 n² — A Step-by-Step Explanation", "---", "Introduction\nMathematical inequalities like 0.05n log₂(n) < 0.01n² may look abstract at first glance, but understanding and solving them is essential in fields such as computer science, algorithm analysis, and operations research. In this article, we break down how to solve this inequality step-by-step, explain its practical significance, and explore where this expression commonly arises. Whether you're a student, researcher, or programmer, mastering this relatively simple log-linear inequality unlocks deeper insights into complexity analysis and optimization.", "---", "### Understanding the Inequality\nWe are solving:\n[ 0.05n \log_2(n) < 0.01n^2 ]", "This compares a linear-logarithmic term on the left to a quadratic term on the right. The goal is to find all positive integer values of ( n ) that satisfy the inequality.", "---", "### Step 1: Simplify the Inequality\nStart by eliminating common factors. Divide both sides by 0.05n (note: ( n > 0 ), so division is valid and preserves inequality direction):\n[ \log_2(n) < \frac{0.01n^2}{0.05n} ]\n[ \log_2(n) < 0.2n ]", "Now the inequality simplifies to:\n[ \log_2(n) < 0.2n ]", "---", "### Step 2: Analyze Behavior and Find Critical Points\nThe inequality involves a logarithmic function growing slowly and a linear function growing faster. To find where equality occurs, solve:\n[ \log_2(n) = 0.2n ]", "This equation has no elementary algebraic solution, so we solve it numerically or graphically. Try small integer values of ( n ):", "- ( n = 1 ): ( \log_2(1) = 0 ), ( 0.2 \ imes 1 = 0.2 ) → ( 0 < 0.2 ) ✅\n- ( n = 4 ): ( \log_2(4) = 2 ), ( 0.2 \ imes 4 = 0.8 ) → ( 2 > 0.8 ) ❌\n- ( n = 8 ): ( \log_2(8) = 3 ), ( 0.2 \ imes 8 = 1.6 ) → ( 3 > 1.6 ) ❌\n- ( n = 16 ): ( \log_2(16) = 4 ), ( 0.2 \ imes 16 = 3.2 ) → ( 4 > 3.2 ) ❌\n- ( n = 32 ): ( \log_2(32) = 5 ), ( 0.2 \ imes 32 = 6.4 ) → ( 5 < 6.4 ) ✅", "So, the inequality flips between ( n = 16 ) and ( n = 32 ). Testing intermediate values shows equality roughly holds near ( n \approx 20 ).", "---", "### Step 3: Confirm Boundaries Numerically\nWe test around the suspected crossover point to pinpoint the exact range.", "Define function:\n[ f(n) = 0.2n - \log_2(n) ]", "Compute ( f(n) ) at key points:\n- ( n = 20 ): ( f(20) = 0.2 \ imes 20 - \log_2(20) \approx 4 - 4.32 = -0.32 ) ❌\n- ( n = 25 ): ( f(25) \approx 5 - 4.64 = 0.36 ) ✅\n- ( n = 15 ): ( f(15) \approx 3 - 3.91 = -0.91 ) ❌", "Actually, since ( f(16) = 3.2 - 4 = -0.8 ), ( f(15) < 0 ), but ( f(16) < 0 ) and ( f(25) > 0 ), the function crosses zero between 16 and 25.", "Using logarithm approximations and iterative numerical methods (like Newton-Raphson), we find the root of ( \log_2(n) = 0.2n ) is approximately ( n \approx 21.9 ).", "Thus, inequality holds when:\n[ 0.05n \leq \log_2(n) < 0.2n \quad \Rightarrow \quad 1 < n < 21.9 ]", "So, the strict inequality holds only for positive integers ( n = 2, 3, ..., 21 ).", "---", "### Step 4: Verify Edge Values\n- At ( n = 1 ): ( 0.05(1)\log_2(1) = 0 < 0.01(1)^2 = 0.01 ) ✅\n- At ( n = 2 ): ( 0.05 \cdot 2 \cdot 1 = 0.1 < 0.04 = 0.01 \cdot 4 ) ✅\n- At ( n = 21 ):\n Left: ( 0.05 \cdot 21 \cdot \log_2(21) \approx 1.05 \cdot 4.392 \approx 4.61 )\n Right: ( 0.01 \cdot 441 = 4.41 ) → ( 4.61 > 4.41 ) ❌\n So ( n = 21 ) does not satisfy the inequality.\n- At ( n = 20 ):\n Left: ( 0.05 \cdot 20 \cdot \log_2(20) \approx 1 \cdot 4.32 \approx 4.32 )\n Right: ( 0.01 \cdot 400 = 4.00 ) → ( 4.32 > 4.00 ) ❌", "Wait — this contradicts earlier bound. Let’s re-analyze:\nWe had ( \log_2(n) < 0.2n ). At ( n=20 ):\n( \log_2(20) \approx 4.32 ), ( 0.2 \ imes 20 = 4.00 ) → ( 4.32 > 4.00 ), so inequality fails.", "But our earlier numeric test with simplified inequality ( \log_2(n) < 0.2n ) at ( n=20 ) gave ( 4.32 < 4.00 ) — incorrect, because ( \log_2(20) \approx 4.32 ), and ( 4 < 4.00 )? No — wait:\nActually, ( \log_2(16) = 4 ), ( \log_2(20) = \log_2(2^2 \cdot 5) = 2 + \log_2(5) \approx 2 + 2.32 = 4.32 ), and ( 0.2 \ imes 20 = 4.00 ), so ( 4.32 > 4.00 ) → inequality fails.", "But earlier numeric check with the original inequality ( 0.05n \log_2(n) < 0.01n^2 ) at ( n=20 ):\nLeft: ( 0.05 \ imes 20 \ imes 4.32 = 4.32 )\nRight: ( 0.01 \ imes 400 = 4.00 ) → ( 4.32 < 4.00 )? ❌ No — 4.32 > 4.00 → inequality fails", "So inequality only holds when ( \log_2(n) < 0.2n ), which occurs only for small ( n ).", "Let’s fix the logic:", "Define:\n[ f(n) = \log_2(n) - 0.2n ]\nWe seek ( f(n) < 0 )", "From earlier:\n- ( n = 1 ): ( 0 - 0.2 = -0.2 < 0 ) ✅\n- ( n = 2 ): ( 1 - 0.4 = 0.6 > 0 ) ❌\n- ( n = 3 ): ( \log_2(3) \approx 1.58 ), ( 0.2 \ imes 3 = 0.6 ), ( 1.58 - 0.6 = 0.98 > 0 ) ❌\n- ( n = 4 ): ( 2 - 0.8 = 1.2 > 0 ) ❌\n- ( n = 8 ): ( 3 - 1.6 = 1.4 > 0 ) ❌\n- ( n = 16 ): ( 4 - 3.2 = 0.8 > 0 ) ❌\n- ( n = 32 ): ( 5 - 6.4 = -1.4 < 0 ) ✅", "But wait — at ( n = 1 ): ( \log_2(1) = 0 < 0.2 \ imes 1 = 0.2 ) → ✅\nIs there any ( n > 1 ) where ( \log_2(n) < 0.2n )?", "Try:\n- ( n = 2 ): ( 1 < 0.4 )? No\n- ( n = 4 ): ( 2 < 0.8 )? No\n- ( n = 8 ): ( 3 < 1.6 )? No\n- ( n = 32 ): ( 5 < 6.4 )? Yes\nBut between 16 and 32? Try ( n = 24 ):\n( \log_2(24) \approx \log_2(16 \cdot 1.5) = 4 + \log_2(1.5) \approx 4 + 0.585 = 4.585 )\n( 0.2 \ imes 24 = 4.8 ) → ( 4.585 < 4.8 ) ✅", "( n = 22 ):\n( \log_2(22) \approx \log_2(16 \cdot 1.375) \approx 4 + \log_2(1.375) \approx 4 + 0.46 = 4.46 )\n( 0.2 \ imes 22 = 4.4 ) → ( 4.46 > 4.4 ) ❌", "( n = 23 ):\n( \log_2(23) \approx \log_2(16 \cdot 1.4375) \approx 4 + 0.52 = 4.52 )\n( 0.2 \ imes 23 = 4.6 ) → ( 4.52 < 4.6 ) ✅", "( n = 21 ):\n( \log_2(21) \approx 4.39 ), ( 0.2 \ imes 21 = 4.2 ) → ( 4.39 > 4.2 ) ❌", "So the solution set is:\nFor ( n = 1 ): ✅\nThen, ( n \geq 23 ) and increasing — solve ( \log_2(n) < 0.2n )", "From above:\n- At ( n = 23 ): ✅\n- At ( n = 22 ): ❌\n- At ( n = 24 ): ✅\nBut check ( n = 22 ): ( \log_2(22) \approx 4.46 > 4.4 ) → no\n( n = 23 ): ( \approx 4.52 < 4.6 ) → yes", "Now, as ( n ) increases, ( 0.2n ) grows linearly, ( \log_2(n) ) grows slower — so inequality holds for all ( n ) from 23 onward until ( \log_2(n) \geq 0.2n ) again? But exponential-log growth never overtakes linear-log, soafter a point it stays true.", "But is there an upper bound? No — for large ( n ), ( 0.2n ) dominates. But from tests, only starting at n = 23 does it hold again.", "Wait — deeper analysis: define ( f(n) = \log_2(n) - 0.2n ). Derivative:\n[ f'(n) = \frac{1}{n \ln 2} - 0.2 ]\nSet ( f'(n) = 0 ):\n[ n = \frac{1}{0.2 \ln 2} \approx \frac{1}{0.1386} \approx 7.22 ]\nSo function increases up to ( n \approx 7.22 ), then decreases toward zero negative. But we care about when ( f(n) < 0 ). Since ( f(n) \ o -\infty ), and positive only near ( n=1 ), it crosses zero once at ( n \approx 2 ), then stays negative until ( n \approx 23 ), then becomes positive and stays positive? Or does it?", "Wait: at ( n = 100 ):\n( \log_2(100) \approx 6.64 ), ( 0.2 \ imes 100 = 20 ) → ( 6.64 < 20 ) → ✅", "So always true for ( n \geq 23 )? Let's confirm monotonicity after peak.", "But from numerical tests:\n- ( n = 22 ): ≈4.46 > 4."]