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

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

Question 1 Report

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

Read lesson note on Modular Arithmetic (WAEC)

Read lesson note on Logarithms (WAEC)

Modular Arithmetic

Logarithms

View Answer

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

Download The App On Google Playstore

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

Related Questions

Download The App On Google Playstore