Given that log4(Y – 1) + log4((frac{1}{2})x) = 1 and log2(y + 1) + log2x…

Given that log₄(Y – 1) + log₄((frac{1}{2})x) = 1 and log₂(y + 1) + log₂x = 2, solve for x and y respectively

Correct Answer: Option C

Explanation

log₄(y – 1) + log₄((frac{1}{2})x) = 1
log₄(y – 1)((frac{1}{2})x) (to) (y – 1)((frac{1}{2})x) = 4 ……..(1)
log₂(y + 1) + log₂x = 2
log₂(y + 1)x = 2 (to) (y + 1)x = 2₂ = 4…..(ii)
From equation (ii) x = (frac{4}{y + 1})……..(iii)
put equation (iii) in (i) = y (y – 1)[(frac{1}{2}(frac{4}{y – 1}))] = 4
= 2y – 2
= 4y + 4
2y = -6
y = -3
x = (frac{4}{-3 + 1})
= (frac{4}{-2})
X = 2
therefore x = -2, y = -3