Question #47488

First observe that $36 = {6}^{2}$.
Now since ${q}^{7} = {q}^{6} \cdot q$, we have
$\sqrt{36 {q}^{7}} = \sqrt{36 {q}^{6}} \cdot \sqrt{q}$
$= 6 {q}^{3} \sqrt{q}$.
Observe that the exponent, 6, is divided by the index of the radical, 2. $\frac{6}{2} = 3$ is the new exponent.