<p>The number of real solutions of the equation \(9^{\log_3(\log_e x)} = \log_e x - (\log_e x)^2 + 1\) is:</p>
Step-by-Step Solution
Key Concept: Simplify the left side using logarithm properties, then substitute t = log_e x to convert into a solvable algebraic equation. Check domain restrictions carefully since logarithms require positive arguments.
<p><strong>Step 1: Simplify the left side.</strong></p><p>We have 9^(log₃(log_e x)). Using the property a^(log_a b) = b, we first rewrite 9 = 3²:</p><p>9^(log₃(log_e x)) = (3²)^(log₃(log_e x)) = 3^(2·log₃(log_e x))</p><p>Using the logarithm property n·log_a b = log_a(b^n):</p><p>3^(log₃((log_e x)²)) = (log_e x)²</p><p><strong>Step 2: Set up the equation.</strong></p><p>The equation becomes: (log_e x)² = log_e x - (log_e x)² + 1</p><p><strong>Step 3: Substitute t = log_e x.</strong></p><p>Let t = log_e x. Then: t² = t - t² + 1</p><p>Simplifying: 2t² - t - 1 = 0</p><p><strong>Step 4: Solve the quadratic.</strong></p><p>Factoring: (2t + 1)(t - 1) = 0</p><p>So t = -1/2 or t = 1</p><p><strong>Step 5: Determine domain restrictions.</strong></p><p>For log₃(log_e x) to be defined, we need log_e x > 0, which means x > 1.</p><p>This requires t = log_e x > 0.</p><p><strong>Step 6: Check solutions against domain.</strong></p><p>• t = -1/2: Since -1/2 < 0, this violates the domain restriction. Not valid.</p><p>• t = 1: Since 1 > 0, this is valid. This gives log_e x = 1, so x = e.</p><p><strong>Step 7: Verify x = e in the original equation.</strong></p><p>LHS: 9^(log₃(1)) = 9^0 = 1</p><p>RHS: 1 - 1 + 1 = 1 ✓</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B