If $x$, $y$, and $z$ are positive real numbers satisfying: \begin{align*}
\log x - \log y &= a, \\
\log y - \log z &= 15, \text{ and} \\
\log z - \log x &= -7, \\
\end{align*}where $a$ is a real number, what is $a$?
I notice that this problem involves logarithms with the same base, so I can use the properties of logarithms to simplify the equations. For example, $\log x - \log y = a$ can be rewritten as $\log \frac{x}{y} = a$, using the quotient rule. Similarly, $\log y - \log z = 15$ can be rewritten as $\log \frac{y}{z} = 15$, and $\log z - \log x = -7$ can be rewritten as $\log \frac{z}{x} = -7$. Now I have three equations with three unknowns, $\log \frac{x}{y}$, $\log \frac{y}{z}$, and $\log \frac{z}{x}$. I wonder if I can eliminate one of them by adding or subtracting the equations. I notice that if I add the first and the second equation, I get $\log \frac{x}{y} + \log \frac{y}{z} = a + 15$. But by the product rule of logarithms, $\log \frac{x}{y} + \log \frac{y}{z} = \log \frac{x}{y} \cdot \frac{y}{z} = \log \frac{x}{z}$. So I have $\log \frac{x}{z} = a + 15$. But wait, I also have $\log \frac{z}{x} = -7$. By the inverse rule of logarithms, $\log \frac{z}{x} = -\log \frac{x}{z}$. So I can substitute that into the equation and get $-\log \frac{x}{z} = -7$. Multiplying both sides by -1, I get $\log \frac{x}{z} = 7$. Now I have two equations that say $\log \frac{x}{z} = 7$ and $\log \frac{x}{z} = a + 15$. This means that $7 = a + 15$, or $a = -8$. 