Positive real numbers a, b, c such that a > 10 ≥ b ≥ c. Given: log₈(a + b) / log₂c + log₂₇(a − b) / log₃c = 2/3 Find the greatest possible integer value of a.
Question:
Positive real numbers a, b, c such that a > 10 ≥ b ≥ c. Given: log₈(a + b) / log₂c + log₂₇(a − b) / log₃c = 2/3 Find the greatest possible integer value of a.
Positive real numbers a, b, c such that a > 10 ≥ b ≥ c. Given: log₈(a + b) / log₂c + log₂₇(a − b) / log₃c = 2/3 Find the greatest possible integer value of a.
Options
Answer: 14
Explanation:
Rewrite logarithms in terms of base c: log₈(a + b) / log₂c = (log(a + b) / log8) ÷ (logc / log2) = (1/3) log_c(a + b) log₂₇(a − b) / log₃c = (log(a − b) / log27) ÷ (logc / log3) = (1/3) log_c(a − b) So the equation becomes: (1/3) log_c(a + b) + (1/3) log_c(a − b) = 2/3 Multiply both sides by 3: log_c(a + b) + log_c(a − b) = 2 Combine logarithms: log_c[(a + b)(a − b)] = 2 (a + b)(a − b) = c² a² − b² = c² a² = b² + c² To maximize a (integer), choose b = c = 10: a² = 10² + 10² = 200 a ≤ √200 ≈ 14.14 Greatest possible integer value of a = 14
Explanation:
Rewrite logarithms in terms of base c: log₈(a + b) / log₂c = (log(a + b) / log8) ÷ (logc / log2) = (1/3) log_c(a + b) log₂₇(a − b) / log₃c = (log(a − b) / log27) ÷ (logc / log3) = (1/3) log_c(a − b) So the equation becomes: (1/3) log_c(a + b) + (1/3) log_c(a − b) = 2/3 Multiply both sides by 3: log_c(a + b) + log_c(a − b) = 2 Combine logarithms: log_c[(a + b)(a − b)] = 2 (a + b)(a − b) = c² a² − b² = c² a² = b² + c² To maximize a (integer), choose b = c = 10: a² = 10² + 10² = 200 a ≤ √200 ≈ 14.14 Greatest possible integer value of a = 14
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