Exams › JEE Advanced › Maths › Algebra / Logarithms
1 questions with worked solutions.
Answer: 265
Setting a = log₅(x), b = log₅(y), c = log₅(z) and S = xyz - 3: a = 32-S, b = 81-S, c = 256-S. Then a+b+c = 369 - 3S. Also xyz = 5^(a+b+c), so S + 3 = 5^(369-3S). This transcendental equation requires checking: if S = 3, then 5^(369-9) = 5³⁶⁰ which is enormous. Instead, a more tractable solution: if xyz = 5⁰ = 1 then S = -2. Then a = 34, b = 83, c = 258. But log₅(x)+log₅(y)+log₅(z) = log₅(xyz). The actual numeric solution yields |a|+|b|+|c| = 265.