Aptitude Logarithm Test Paper 11) Find the logarithm of 1/256 to the base 2√2.
Answer: C Explanation: Let log_{2√2} [1/256] = x We know that log_{a} y = x is similar to a^{x} = y So, we can write it as [1/256] = (2√2) ^{x} Or, (2√2) ^{x} = [1/2^{8}] Or, [2^{1} * 2^{1/2}]^{x} = 1/2^{8} Or, 2^{3x/2} = 28 Therefore, 3x/2 = 8 Hence, x = (8 * 2)/ 3 = 16/3 2) If log_{a} [1/36] = 2/3, find the value of a.
Answer: D Explanation: ATQ, log_{a} [1/36] = 2/3 We know that log_{a} y = x is similar to a^{x} = y So, a ^{ (2/3) } = 1/36 Or, a ^{ (2/3) } = 1/6^{2} Or, a^{ (2/3) }= 6^{2} Now, multiply and divide by 3 in the power of 6 to make the power equals to power of a. So, a ^{ (2/3) } = 6^{3(2/3) } On comparing both side, we get a = 6^{3} Therefore, a = 216 3) Find the value of x Log_{4}(log_{8} 64) = log_{5} x
Answer: D Explanation: ATQ, Log_{4} (log_{8} 64) = log_{5} x...... (i) Let, log_{8} 64 = a Or, 64 = 8^{a}, or 8^{a} = 8^{2} That means a=2 Now, put log_{8} 64 = 2in equation 1. Log_{4} (2) = log_{5} x...... (ii) Now, let log_{4} 2 = s Or, 4^{s} = 2, or 2^{2S} = 2 Now, on comparing both side we get 2_{s} = 1, or s = ½ Put the value of s in equation 2 So, log_{5} x = ½ Therefore, x = 5^{1/2} = √5 4) The equation, a^{2} + b^{2} = 7ab equals to Answer: A Explanation: Here, a^{2} + b^{2} = 7ab Add both sides 2ab to make a formula. Or, a^{2} + b^{2} + 2ab = 7ab + 2ab Or, (a+b) ^{2} = 9ab Or, (a+b) ^{2} / 9 = ab Or, [1/3 (a+b)]^{2} = ab Now, taking log both sides Therefore, log [1/3 (a+b)]^{2} = log ab We know that log m*n = log m + log n, and log m^{n} = n log m So, 2 log [1/3 (a+b)] = log a + log b Therefore, log [1/3 (a+b)] = 1/2(log a + log b) = a^{2} + b^{2} = 7ab 5) If (log_{3} x)(log_{x} 2x)(log_{2x} y) = log_{x} x^{2}, find y.
Answer: A Explanation: (log_{3} x)(log_{x} 2x)(log_{2x} y) = log_{x} x^{2} B = C = Now, ATQ, A* B * C = log_{x} x^{2} We know that, log_{a} m^{n} = n log_{a} m We know that log_{a} a = 1 Aptitude Logarithm Test Paper 2 Aptitude Logarithm Test Paper 3 Aptitude Logarithm Test Paper 4
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