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

Mathematics

JAMB 1998

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

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Correct Answer: Option 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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