Given that log4(y - 1) + log4(12 1 2 x) = 1 and log2(y + 1) + log2x = 2, solve for x and y respectively

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

Answer Details

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