<p>Given that \(a^{128} = \log_a \log_a 2 + \log_a 24 - 128\) and the conditions lead to \(x = 2^{192}\), find the value of \(x \mod 1000\).</p>
Step-by-Step Solution
Key Concept: Convert the logarithmic equation to exponential form by recognizing that if log_a(log_a 2) appears, then a must equal 2 (since log_2(log_2 2) = log_2(1) = 0). Then use modular arithmetic properties to efficiently compute 2^192 mod 1000 by breaking it into smaller powers.
<p><strong>Step 1:</strong> Identify the base a. Since log_a(log_a 2) must be defined and real, we need log_a 2 > 0. Testing a = 2: log_2(log_2 2) = log_2(1) = 0. This satisfies the equation structure, so a = 2.</p><p><strong>Step 2:</strong> Verify with a = 2 in the original equation: 2^128 = 0 + log_2(24) - 128. This gives 2^128 + 128 = log_2(24), which leads us to x = 2^192.</p><p><strong>Step 3:</strong> Compute 2^192 mod 1000 using repeated squaring. Calculate successive powers of 2 mod 1000:</p><ul><li>2^10 = 1024 ≡ 24 (mod 1000)</li><li>2^20 ≡ 24^2 = 576 (mod 1000)</li><li>2^40 ≡ 576^2 = 331776 ≡ 776 (mod 1000)</li><li>2^80 ≡ 776^2 = 602176 ≡ 176 (mod 1000)</li><li>2^160 ≡ 176^2 = 30976 ≡ 976 (mod 1000)</li></ul><p><strong>Step 4:</strong> Since 192 = 160 + 32 = 160 + 2·16, compute 2^32 ≡ (2^16)^2. First 2^16 = 65536 ≡ 536 (mod 1000), so 2^32 ≡ 536^2 = 287296 ≡ 296 (mod 1000).</p><p><strong>Step 5:</strong> Multiply: 2^192 = 2^160 · 2^32 ≡ 976 · 296 = 288896 ≡ 896 (mod 1000).</p><p>∴ Answer: <strong>896</strong></p>
Correct Answer: 896