Question #291933

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





1
Expert's answer
2022-01-31T16:06:49-0500

(y1)(x)=4(y-1)(x)=4 ......(i)......(i)

(y+1)(x)=22.....(ii)(y+1)(x)=2^2.....(ii)


Dividing (ii)(ii) by (i)(i)


y1y+1=1\frac{y-1}{y+1}=1

y1=y+1y-1=y+1

0=20=2


\therefore No solution


Rewriting the question to read ;


log4(y1)+_4(y-1)+ log4(12x)=1_4(\frac{1}{2}x)=1

log2(y+1)+_2(y+1)+ log2x=2_2x=2


Then : Using laws of logarithms

12x(y1)=4....(i)\frac{1}{2}x(y-1)=4....(i)

x(y+1)=4.....(ii)x(y+1)=4.....(ii)


Dividing (ii)(ii) by (i)(i)

2(y+1)y1=1\frac{2(y+1)}{y-1}=1


Solving

y=3y=-3


Substituting in (ii)(ii)

x(3+1)=4x(-3+1)=4

x=2x=-2










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