$\begin{align}  & (i)x+y-z={{\log }_{a}}\left( \frac{a}{a+1} \right)+{{\log }_{a}}\left( \frac{a}{a-1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & x+y-z={{\log }_{a}}\left( \frac{a}{a+1} \right)\left( \frac{a}{a+1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & x+y-z={{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & x+y-z={{\log }_{a}}\frac{\frac{{{a}^{2}}}{{{a}^{2}}-1}}{\frac{{{a}^{2}}}{{{a}^{2}}-1}}={{\log }_{a}}1=0 \\ & (ii)\text{ }x+2y-z={{\log }_{a}}\left( \frac{a}{a+1} \right)+2{{\log }_{a}}\left( \frac{a}{a-1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & x+2y-z={{\log }_{a}}\left( \frac{a}{a+1} \right)+{{\log }_{a}}{{\left( \frac{a}{a-1} \right)}^{2}}-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & x+2y-z={{\log }_{a}}\left( \frac{a}{a+1} \right)\left( \frac{{{a}^{2}}}{{{(a-1)}^{2}}} \right)-{{\log }_{a}}\frac{{{a}^{2}}}{(a-1)(a+1)} \\ & x+2y-z={{\log }_{a}}\frac{{{a}^{3}}}{(a+1){{(a-1)}^{2}}}-{{\log }_{a}}\frac{{{a}^{2}}}{(a-1)(a+1)} \\ & x+2y-z={{\log }_{a}}\frac{\tfrac{{{a}^{3}}}{(a+1){{(a-1)}^{2}}}}{\tfrac{{{a}^{2}}}{(a-1)(a+1)}} \\ & x+2y-z={{\log }_{a}}\left[ \frac{{{a}^{3}}}{(a+1){{(a-1)}^{2}}}\times \frac{(a-1)(a+1)}{{{a}^{2}}} \right] \\ & x+2y-z={{\log }_{a}}\left( \frac{a}{a-1} \right) \\ & Alternative\text{ }approach\text{ }x+2y-z=(x+y-z)+y\text{ } \\ & But\text{ }x+y-z=0\text{ }(i.e\text{ }question\text{ }i) \\ & x+2y-z=(x+y-z)+y=0+y={{\log }_{a}}\left( \frac{a}{a-1} \right) \\ & x+2y-z={{\log }_{a}}\left( \frac{a}{a-1} \right) \\ &  \\ & (iii)\text{ } \\ & z-x-y={{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right)-{{\log }_{a}}\left( \frac{a}{a+1} \right)-{{\log }_{a}}\left( \frac{a}{a-1} \right) \\ & z-x-y={{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right)-\left[ {{\log }_{a}}\left( \frac{a}{a+1} \right)+{{\log }_{a}}\left( \frac{a}{a-1} \right) \right] \\ & z-x-y={{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right)-{{\log }_{a}}\left( \frac{a}{a+1} \right)\left( \frac{a}{a-1} \right) \\ & z-x-y={{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right)=0 \\ & Alternative\text{ }approach\text{ }z-x-y=-(x+y-z) \\ & From\text{ }question\text{ }(i)\text{ }x+y-z=0 \\ &  \\ & (iv)\text{ }2x+y-z=2{{\log }_{a}}\left( \frac{a}{a+1} \right)+{{\log }_{a}}\left( \frac{a}{a-1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & 2x+y-z={{\log }_{a}}{{\left( \frac{a}{a+1} \right)}^{2}}+{{\log }_{a}}\left( \frac{a}{a-1} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & 2x+y-z={{\log }_{a}}\left[ {{\left( \frac{a}{a+1} \right)}^{2}}\times \frac{a}{a-1} \right]-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & 2x+y-z={{\log }_{a}}\left( \frac{{{a}^{3}}}{(a+1)({{a}^{2}}-1)} \right)-{{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & 2x+y-z={{\log }_{a}}\left( \frac{{{a}^{3}}}{(a+1)({{a}^{2}}-1)}\times \frac{{{a}^{2}}-1}{a} \right) \\ & 2x+y-z={{\log }_{a}}\frac{a}{a+1} \\\end{align}$