Given that \( \log_{4}(y - 1) + \log_{4}\left(\frac{1}{2}x\right) = 1 \) and \( \log_{2}(y + 1) + \log_{2}x = 2 \), solve for x and y respectively
Question 1 Report
Given that \( \log_{4}(y - 1) + \log_{4}\left(\frac{1}{2}x\right) = 1 \) and \( \log_{2}(y + 1) + \log_{2}x = 2 \), solve for x and y respectively
Answer Details
log4(y - 1) + log4(x) = 1
log4(y - 1)(x) (y - 1)(x) = 4 ........(1)
log2(y + 1) + log2x = 2
log2(y + 1)x = 2 (y + 1)x = 22 = 4.....(ii)
From equation (ii) x = ........(iii)
put equation (iii) in (i) = y (y - 1)[)] = 4
= 2y - 2
= 4y + 4
2y = -6
y = -3
x =
=
X = 2
therefore x = -2, y = -3
log4(y - 1)(x) (y - 1)(x) = 4 ........(1)
log2(y + 1) + log2x = 2
log2(y + 1)x = 2 (y + 1)x = 22 = 4.....(ii)
From equation (ii) x = ........(iii)
put equation (iii) in (i) = y (y - 1)[)] = 4
= 2y - 2
= 4y + 4
2y = -6
y = -3
x =
=
X = 2
therefore x = -2, y = -3
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