**Given that log**_{4}(y - 1) + log_{4}(\(\frac{1}{2}\)x) = 1 and log_{2}(y + 1) + log_{2}x = 2,...

_{4}(y - 1) + log

_{4}(\(\frac{1}{2}\)x) = 1 and log

_{2}(y + 1) + log

_{2}x = 2,...

### Question

Given that log_{4}(y - 1) + log_{4}(\(\frac{1}{2}\)x) = 1 and log_{2}(y + 1) + log_{2}x = 2, solve for x and y respectively

### Options

A) 2, 3

B) 3, 2

C) -2, -3

D) -3, -2

The correct answer is C.

### Explanation:

log

log

log

log

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

_{4}(y - 1) + log_{4}(\(\frac{1}{2}\)x) = 1log

_{4}(y - 1)(\(\frac{1}{2}\)x) \(\to\) (y - 1)(\(\frac{1}{2}\)x) = 4 ........(1)log

_{2}(y + 1) + log_{2}x = 2log

_{2}(y + 1)x = 2 \(\to\) (y + 1)x = 2_{2}= 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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log

_{4}(y - 1) + log_{4}(\(\frac{1}{2}\)x) = 1log

_{4}(y - 1)(\(\frac{1}{2}\)x) \(\to\) (y - 1)(\(\frac{1}{2}\)x) = 4 ........(1)log

_{2}(y + 1) + log_{2}x = 2log

_{2}(y + 1)x = 2 \(\to\) (y + 1)x = 2_{2}= 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