Exams › JEE Advanced › Maths

Evaluate the product: log₂(4) * log₄(5) * log₅(10) * log₁₀(32).

1. 2
2. 5
3. 4
4. 3

Correct answer: 5

Solution

By the chain (change-of-base) rule, the product log₂(4)*log₄(5)*log₅(10)*log₁₀(32) telescopes to log₂(32) = 5.

Related JEE Advanced Maths questions

• Given log₅(N) = I1 + f1 and log₃(N) = I2 + f2, where I1 and I2 are non-negative integers and f1, f2 are in [0, 1). If I1 = 3 and I2 = 4, find the number of possible integral values of N.
• Let m and n be positive real numbers such that m^(log₃ 5) = 9 and n^(log₅ 7) = 25. Also (log₃ m)*(log₅ n) = log₈ a, where b is a prime number. Match List I with List II. List I: (P) m^((log₃ 5)²) + n^((log₅ 7)²) (Q) value of a (R) value of b (S) antilog_(sqrt(b))(log_b 2401) List II: (1) 81 (2) 7 (3) 49 (4) 74
• If logₓ(a) = b for permissible values of a and x, which of the following statements can be correct?
• Let M be the number of digits in 11⁵⁰ and N be the number of zeros after the decimal point but before the first significant digit in (0.11)⁵⁰. Given that log10(11) = 1.041, identify the correct statement(s).
• Given log₁₀(2) = 0.3010 and log₁₀(3) = 0.4771, let P be the number of digits in 4⁸ * 3⁵ * 5³. Find the value of (P - 1) / 2.
• Find the absolute value of log base 0.001 of 1000.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →