Home » » Given that log4(y - 1) + log4($$\frac{1}{2}$$x) = 1 and log2(y + 1) + log2x = 2,...

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

Question

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

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

Dicussion (1)

• 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