Quantitative Aptitude – Algebra – Logarithm

Slot -1 – Quantitative Aptitude – Algebra – Logarithm – Let x and y be positive real numbers such that

Q. Let x and y be positive real numbers such that log(base 5) (x + y) + log(base 5) (x − y) = 3, and log(base 2)y − log(base 2)x = 1 − log(base 2)3. Then xy equals?

1. 150
2. 100
3. 25
4. 250

Answer: 150

Solution: Given, log(base5) (x + y) + log(base5) (x − y) = 3
Or log(base5) (x + y)*(x-y) =3
Or x^2 –y^2 = 5^3 = 125————-1)
log(base2) y − log(base2) x = 1 − log(base2) 3= log(base2) 2 – log(base2) 3
log(base2) y/x = log(base2) 2/3
y/x = 2/3
y = 2x/3
from eq 1) x^2 – (2x/3)^2 = 125
x^2 – (4x^2/9) = 125
5x^2 = 125*9 or x^2 = 225
x = 15
y= 2x/3 = 30/3 = 10
xy = 15*10 =150

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