Home » 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…

Given that log4(Y – 1) + log4((frac{1}{2})x) = 1 and log2(y + 1) + log2x = 2, solve for x and y respectively
  • A.
    2, 3
  • B.
    3, 2
  • C.
    -2, -3
  • D.
    -3, -2
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

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