Discussion
If \(pqr \neq 0\) and \(p^{-x} = \frac{1}{q}, q^{-y} = \frac{1}{r}, r^{-z} = \frac{1}{p}\), what is the value of the product \(xyz\)?
Answer: C
The product is \(xyz = 1\).
Taking reciprocals, \(p^{-x} = \frac{1}{q}\) gives \(p^{x} = q\); likewise \(q^{y} = r\) and \(r^{z} = p\). Substitute each into the next:
\[ \begin{aligned} r &= q^{y} = (p^{x})^{y} = p^{xy} \\ p &= r^{z} = p^{xyz} \end{aligned} \]
So \(p^{xyz} = p^{1}\), and \(xyz = 1\) for a general base \(p\). The answer \(-1\) comes from forgetting that the negative exponents were cleared by the reciprocals, and \(\frac{1}{pqr}\) and \(pqr\) mix the bases into the exponents, which the equations never do.
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