Answer to Question #291933 in Algebra for Rossy

Question

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

Answer 1

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

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

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

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

$y-1=y+1$

$0=2$

$\therefore$ No solution

Rewriting the question to read ;

log$_4(y-1)+$ log$_4(\frac{1}{2}x)=1$

log$_2(y+1)+$ log$_2x=2$

Then : Using laws of logarithms

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

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

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

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

Solving

$y=-3$

Substituting in $(ii)$

$x(-3+1)=4$

$x=-2$