(y−1)(x)=4 ......(i)
(y+1)(x)=22.....(ii)
Dividing (ii) by (i)
y+1y−1=1
y−1=y+1
0=2
∴ No solution
Rewriting the question to read ;
log4(y−1)+ log4(21x)=1
log2(y+1)+ log2x=2
Then : Using laws of logarithms
21x(y−1)=4....(i)
x(y+1)=4.....(ii)
Dividing (ii) by (i)
y−12(y+1)=1
Solving
y=−3
Substituting in (ii)
x(−3+1)=4
x=−2
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