Find smallest n such that 64000 / 2^n = 1 → 2^n = 64000 → n = log₂(64000)

["# Find the Smallest Integer ( n ) Such That ( \frac{64000}{2^n} = 1 ): A Clear Step-by-Step Explanation", "When solving equations involving exponents, understanding logarithms is essential—especially when finding the smallest integer ( n ) such that an expression equals 1. In this article, we’ll explore how to determine the smallest integer ( n ) satisfying:", "[\n\frac{64000}{2^n} = 1\n]", "which leads to solving:", "[\n2^n = 64000\n]", "Then, we find:", "[\nn = \log_2(64000)\n]", "This guide breaks down the process clearly and helps you apply logarithms efficiently in similar problems.", "---", "## Why Use Logarithms Here?", "The equation ( 2^n = 64000 ) is exponential. To isolate ( n ), we naturally use logarithms—specifically, base-2 logarithms—because 64000 is a power-of-2 candidate (or close to it). Recall that:", "[\nn = \log_2(64000)\n]", "This gives an exact expression for ( n ). But since the question asks for the smallest integer ( n ), we must evaluate this logarithm and round appropriately.", "---", "## Step 1: Prime Factorization of 64000", "To compute ( \log_2(64000) ), it helps to express 64000 as a product of prime powers.", "Break down 64000:", "[\n64000 = 64 \ imes 1000 = 2^6 \ imes (10^3) = 2^6 \ imes (2 \ imes 5)^3 = 2^6 \ imes 2^3 \ imes 5^3\n]", "Now combine powers of 2:", "[\n64000 = 2^{6+3} \ imes 5^3 = 2^9 \ imes 5^3\n]", "---", "## Step 2: Compute ( \log_2(64000) )", "Using properties of logarithms:", "[\n\log_2(64000) = \log_2(2^9 \ imes 5^3) = \log_2(2^9) + \log_2(5^3) = 9 + 3\log_2(5)\n]", "Now, estimate ( \log_2(5) ). Recall:", "[\n\log_2(5) = \frac{\ln 5}{\ln 2} \approx \frac{1.6094}{0.6931} \approx 2.3219\n]", "Thus:", "[\nn = 9 + 3 \ imes 2.3219 = 9 + 6.9657 = 15.9657\n]", "---", "## Step 3: Find the Smallest Integer ( n ) Satisfying ( 2^n = 64000 )", "Since ( n ) must be an integer and ( 2^n = 64000 ), ( n ) must be the smallest integer greater than or equal to ( 15.9657 )—that is:", "[\nn = \lceil 15.9657 \rceil = 16\n]", "But wait: does ( 2^{16} = 64000 )? Let’s verify:", "[\n2^{16} = 65536\n]", "This is greater than 64000.", "Try ( 2^{15} ):", "[\n2^{15} = 32768\n]", "Much less than 64000.", "So ( 2^n = 64000 ) has no integer solution—but the original equation asks for the smallest ( n ) such that:", "[\n\frac{64000}{2^n} = 1 \quad \Rightarrow \quad 2^n = 64000\n]", "Since ( 2^{15} = 32768 ) and ( 2^{16} = 65536 ), no integer power of 2 equals exactly 64000. However, the problem can be interpreted as asking:", "> Find the smallest integer ( n ) such that ( \frac{64000}{2^n} \leq 1 ), or equivalently ( 2^n \geq 64000 ).", "Indeed, since ( 2^{16} = 65536 > 64000 ) and ( 2^{15} = 32768 < 64000 ), the smallest ( n ) satisfying ( 2^n \geq 64000 ) is ( n = 16 ), ensuring the fraction is at most 1.", "Hence, the smallest integer ( n ) such that:", "[\n\frac{64000}{2^n} \leq 1\n]", "is:", "[\nn = 16\n]", "---", "## Final Answer", "[\n\boxed{n = 16}\n]", "This value ensures ( \frac{64000}{2^{16}} = \frac{64000}{65536} \approx 0.976 < 1 ), the smallest such ( n ) where the quotient is less than or equal to 1.", "---", "## Summary", "- Start with ( \frac{64000}{2^n} = 1 \Rightarrow 2^n = 64000 )\n- Prime factorization: ( 64000 = 2^9 \ imes 5^3 )\n- So ( n = \log_2(64000) = 9 + 3\log_2(5) \approx 15.9657 )\n- The smallest integer ( n ) where ( 2^n \geq 64000 ) is ( n = 16 )", "Understanding logarithms and prime factorizations allows efficient problem-solving in exponential equations—especially when precise integer values are required.", "---", "Keywords:\nlog₂(64000), smallest integer n, 2^n = 64000, solve exponential equation, logarithms explained, mathematical problem solving, log base 2, powers of 2, decimal to logarithm conversion"]