Question

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

• A. $y=\frac{2xz}{x+z}$
• B. $y^2=xz$
• C. $2y=x+z$
• D. $\frac{x+y}{y-z}=\frac{x}{z}$

Answer 1

The correct option is A $y=\frac{2xz}{x+z}$

Given,
$\ln (x+z)+\ln (x-2y+z)=2\ln (x-z)$

For the given equation to be valid we gave to assume $x+z,x-2y+z,x-z>0$

We have,

$\ln (x+z)+\ln (x-2y+z)=2\ln (x-z)$

$\Rightarrow \ln (x-2y+z)={\ln (x-z)}^2-\ln (x+z),(\because a\log x=\log x^a)$

$\Rightarrow \ln (x-2y+z)=\ln \frac{{(x-z)}^2}{(x+z)},(\because \log a-\log b=\log \frac{a}{b})$

$\Rightarrow x-2y+z=\frac{{(x-z)}^2}{(x+z)},(\because \log a=\log b\iff a=b)$

$\Rightarrow x+z-\frac{{(x-z)}^2}{(x+z)}=2y$
$\Rightarrow \frac{(x+z{)}^2-{(x-z)}^2}{(x+z)}=2y$

$\Rightarrow \frac{4xz}{x+z}=2y$
$\Rightarrow \frac{2xz}{x+z}=y$

$\therefore y=\frac{2xz}{x+z}$