# Quantitative Aptitude – Algebra – Logarithms – If log (2^a × 3^b × 5^c)

January 1st, 2020 by

Quantitative Aptitude - Algebra - Logarithms - If log (2^a × 3^b × 5^c) Quantitative Aptitude - Algebra - Logarithms Question If log (2^a × 3^b × 5^c) is the arithmetic mean of log (2^2 × 3^3 × 5), log (2^6 × 3 × 5^7), and log(2 × 3^2 × 5^4), then a equals Answer 3 Solution From CAT 2017 - Quantitative Aptitude - Algebra - Logarithms, we can see that, log (2^a. 3^b. 5^c) = [log (2^2.3^3.5) + log (2^6.3.5^7) + log (2.3^2.5^4)]/3 3 * log (2^a. 3^b. 5^c) = log (2^9.3^6.5^12) log (2^a. 3^b. 5^c)^3 = log (2^9.3^6.5^12) log (2^3a.

# Quantitative Aptitude – Algebra – Logarithms – If x is a real number

January 1st, 2020 by

Quantitative Aptitude - Algebra - Logarithms - If x is a real number Quantitative Aptitude - Algebra - Logarithms Question If x is a real number such that log(base 3)5 = log(base 5)(2 + x), then which of the following is true? A) 0 < x < 3 B) 23 < x < 30 C) x > 30 D) 3 < x < 23 Answer Option (D) Solution From CAT 2017 - Quantitative Aptitude - Algebra - Logarithms, we can see that, Log(base 3)5 lies between 1 and 2 because Log(base 3)3 = 1 and Log(base 3)9 = 2 1 < Log(base 3)5 < 2 So, log(base 5)(2+x

# Quantitative Aptitude – Algebra – Logarithms – If G is the geometric mean

December 30th, 2019 by

Quantitative Aptitude - Algebra - Logarithms - If G is the geometric mean Quantitative Aptitude - Algebra - Logarithms Question Suppose, log(base3)x = log(base12)y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log(base6)G is equal to A) √a B) 2a C) a/2 D) a Answer Option (D) Solution As per CAT 2017 - Quantitative Aptitude - Algebra - Logarithms, we can see that x=3^a and y=12^a G = √(3^a * 12^a) = 6^a Log (base 6) 6^a = a Option (D) Logarithm Concepts Questions and Answers for

# Quantitative Aptitude – Algebra – Logarithms – 1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100

December 23rd, 2019 by

Quantitative Aptitude – Algebra - Logarithms – 1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100 Slot -2 – Quantitative Aptitude – Algebra - Logarithms – 1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100 1/log(base2⁡)100 -1/log(base4)⁡100 +1/log(base5)⁡100 -1/log(base10)⁡100 +1/log(base20⁡)100 -1/log(base25)⁡100 +1/log(base50)⁡100 ---? a) ½ b) 10 c) -4 d) 0 Answer: a) ½ Solution: Using log(basea)⁡b = 1/log(baseb⁡)a 1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base

# Quantitative Aptitude – Algebra – Logarithms – If p^3 = q^4 = r^5 = s^6

December 23rd, 2019 by

Quantitative Aptitude – Algebra - Logarithms – If p^3 = q^4 = r^5 = s^6 Slot -2 – Quantitative Aptitude – Algebra - Logarithms – If p^3 = q^4 = r^5 = s^6 If p^3 = q^4 = r^5 = s^6, then the value of log_s⁡pqr is equal to? a) 24/5 b) 16/5 c) 47/10 d) 1 Answer: c) 47/10 Solution: Let p^3 = q^4 = r^5 = s^6=k So p=k^(1/3), q=k^(1/4), r=k^(1/5) and s=k^(1/6) Thus log(base s)⁡pqr = log(base(k^(1/6) ))⁡ k^(1/3+1/4+1/5) =6 log(base k)⁡ k^((20+15+12)/60)=6×47/60 log(base k)⁡ k=47/10 Other p

# Quantitative Aptitude – Algebra – Logarithms – If x is a positive quantity such that 2^x

December 21st, 2019 by

Quantitative Aptitude – Algebra - Logarithms – If x is a positive quantity such that 2^x Slot -1 – Quantitative Aptitude – Algebra - Logarithms – If x is a positive quantity such that 2^x If x is a positive quantity such that 2^x = 3^log(base5)^⁡2 , then x is equal to? a) log(base5⁡)^9 b) 1 + log(base5)⁡ 3/5 c) log(base5⁡)^8 d) 1 + log(base3) ⁡5/3 Answer: b) 1 + log(base5)⁡ 3/5 Solution: Given , 2^x = 3^log(base5)⁡2 taking log of both sides , x log⁡2 = log(base5) 2 log⁡ 3 = (log⁡ 2

# Quantitative Aptitude – Algebra – Logarithms – If log(base12)⁡ 81=p then 3

December 21st, 2019 by

Quantitative Aptitude – Algebra - Logarithms – If log(base12)⁡ 81=p then 3 Slot -1 – Quantitative Aptitude – Algebra - Logarithms – If log(base12)⁡ 81=p then 3 If log(base12)⁡ 81=p then 3 { (4-p)/(4+p)} is equal to a) log (base2) 8 b) log (base4) 16 c) log (base6) 8 d) log (base6) 16 Solution: Given , log(base12)⁡ 81 = p 4 log (base12) 3 = p So 3 (4-p)/(4+p) = 3 (1- log(base12)⁡ 3 )/(1+ log(base12)⁡ 3)=3 × log(base12)⁡(12/3)/log(base12)⁡(12×3) = 3×log(base36)⁡ 4 = log(base6)