Question

If $\log (x+z)+\log (x-2y+z)=2\log (x-z)$, then

• A. $y(x+z)=2xz$
• B. $z(z+y)=2xy$
• C. $x+z=3y$
• D. $x^2-z^2=y$

Answer 1

The correct option is A $y(x+z)=2xz$

$\log (x+z)+\log (x-2y+z)=2\log (x-z)\Rightarrow \log (x+z)(x-2y+z)=2\log (x-z)\Rightarrow \log (x+z)(x-2y+z)-\log (x-z)=\log (x-z)\Rightarrow \log \frac{(x+z)(x-2y+z)}{(x-z)}=\log (x-z)\Rightarrow \frac{(x+z)(x-2y+z)}{(x-z)}=(x-z)\Rightarrow (x+z)(x-2y+z)=(x-z{)}^2\Rightarrow (x+z{)}^2-2y(x+z)=(x-z{)}^2\Rightarrow 2xz-2yx-2yz=-2xz\Rightarrow 2xz=xy+yz$