### Instruction: 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 $a$?### Response: I notice that the equations involve logarithms of the same base, which I assume is 10 since it is not specified.
I also notice that each equation is of the form $\log u - \log v = w$, which I can rewrite as $\log (u/v) = w$ by the logarithm quotient rule.
This might make it easier to eliminate some of the variables and solve for $a$.
For example, from the first equation, I can write $\log (x/y) = a$.
Similarly, from the second equation, I can write $\log (y/z) = 15$.
And from the third equation, I can write $\log (z/x) = -7$.
Now I have three equations with three unknowns, but they are all in terms of logarithms.
To get rid of the logarithms, I can exponentiate both sides of each equation using the base 10.
This gives me: \begin{align*}
x/y &= 10^a, \\
y/z &= 10^{15}, \text{ and} \\
z/x &= 10^{-7}.
\\
\end{align*}
Now I have three equations with three unknowns, but they are all in terms of ratios of the variables.
I wonder if I can use these equations to find a product or quotient of the variables that simplifies to a constant.
For example, if I multiply the first equation by the second equation, I get: \begin{align*}
(x/y) \times (y/z) &= 10^a \times 10^{15} \\
\implies x/z &= 10^{a + 15}.
\\
\end{align*}
Similarly, if I multiply the second equation by the third equation, I get: \begin{align*}
(y/z) \times (z/x) &= 10^{15} \times 10^{-7} \\
\implies y/x &= 10^{8}.
\\
\end{align*}
And if I multiply the first equation by the third equation, I get: \begin{align*}
(x/y) \times (z/x) &= 10^a \times 10^{-7} \\
\implies z/y &= 10^{a - 7}.
\\
\end{align*}
Now I have three equations with three unknowns, but they are all in terms of the reciprocals of the variables.
I can rewrite them as: \begin{align*}
x/z &= 10^{a + 15}, \\
z/y &= 10^{a - 7}, \text{ and} \\
y/x &= 10^8.
\\
\end{align*}
I notice that if I multiply all three equations together, I get: \begin{align*}
(x/z) \times (z/y) \times (y/x) &= 10^{a + 15} \times 10^{a - 7} \times 10^8 \\
\implies 1 &= 10^{2a + 16} \\
\end{align*}
This means that $2a + 16 = 0$, or $a = -8$.
Therefore, I have found the value of $a$ that satisfies the system of equations.