Explanation

(b)
If $x,\ y,\ \mathcal{z}$ are in HP, then
$y = \frac{2x\mathcal{z}}{x + \mathcal{z}}$ …(i)
Now, $\log{\left( x + \mathcal{z} \right) + \log{(x - 2y + \mathcal{z)}}\ }$
$= \log{\lbrack(}x + \mathcal{z}$)($\ x - 2y + \mathcal{z}$)]
$= \log\left\lbrack (x + \mathcal{z)}\left( x + \mathcal{z -}\frac{4x\mathcal{z}}{x + \mathcal{z}} \right) \right\rbrack$ [from Eq.(i)]
$= \log\left\lbrack \left( x + \mathcal{z} \right)^{2} - 4x\mathcal{z} \right\rbrack$
$= \log\left( x - \mathcal{z} \right)^{2}$
$= 2\log{(x - \mathcal{z)}}$