jmc

Notice that by the logarithmic identity $\log (x)-\log (y)=\log \frac{x}{y}$, the equations are equivalent to $\log \frac{x}{y}=a$, $\log \frac{y}{z}=15$, and $\log \frac{z}{x}=-7$ respectively. Adding all three equations together yields $\log \frac{x}{y}+\log \frac{y}{z}+\log \frac{z}{x}=a+15-7$. From the identity $\log (x)+\log (y)=\log (xy)$, we obtain $\log \left(\frac{x}{y}\cdot \frac{y}{z}\cdot \frac{z}{x}\right)=a+8$. Cancelations result in $\log (1)=a+8$. Since $\log (1)=0$, we find $a=\boxed{-8}$.