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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

Options

A) 2, 3

B) 3, 2

C) -2, -3

D) -3, -2

The correct answer is 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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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

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