Given that \(p=24\) and \(q=-16\), simplify

\(\dfrac{x^px^q}{x^7}\),

\((x^2)^p(x^3)^q\).

There are three identities we need for this question, namely \[ x^ax^b = x^{a+b},\qquad \frac{x^a}{x^b} = x^{a-b},\qquad \left( x^a \right)^b = x^{ab}. \]

Applying them to \((a)\) we find

\[\begin{align*} \frac{x^px^q}{x^7}&=\frac{x^{p+q}}{x^7}\\ &=x^{p+q-7}\\ &=x^{1}=x. \end{align*}\]For \((b)\) we have

\[\begin{align*} \left(x^2 \right)^p \left(x^3 \right)^q &= x^{2p}x^{3q}\\ &=x^{2p+3q} \\ &= x^0=1. \end{align*}\]