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

Assessment: JAMB UTME - Mathematics - 1998 Subject: General Mathematics

Question 1 Report

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

2, 3

3, 2

-2, -3

-3, -2

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

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

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