2016/014) if $x,y,z$ are in H.P then show that $log(x+z)+ log (x+z-2y) = 2log(x-z)$

We have $x,y,z$ are in HP
so $\dfrac{1}{x} + \dfrac{1}{z} = 2\dfrac{1}{y}$
or $ y(x+z) = 2xz$
Now $(x+z)(x+z-2y) = (x+z) ( x + z - \dfrac{4xz}{x+z})$
$= (x+z)^2 - 4xz$
$= x^2 + z^2 + 2xz -4xz$
$= x^2 + z^2 -2xz = (x-z)^2$
taking log of both sides we get the result