Solving $\frac{\sqrt{108x^{10}}}{\sqrt{2x}}$

Question

Simplify $$\frac{\sqrt{108x^{10}}}{\sqrt{2x}}$$

$\dfrac{\sqrt{108x^{10}}}{\sqrt{2x}}= \dfrac{(108x^{10})^{1/2}}{(2x)^{1/2}}$

The $1/2$ exponent cancels $\implies \dfrac{108x^{10}}{2x}$

$\implies 54x^9$

I was a little confused about the quotient rule for square roots. Is this correct? If not, please show correct version. Thanks.

Answer 1

$$\frac{\sqrt{108x^{10}}}{\sqrt{2x}}= \sqrt{\frac{108x^{10}}{2x}}=\sqrt{54x^9}=3x^4\sqrt{6x}$$

Answer 2

This is incorrect. Generally speaking, we have that $\frac{a^c}{b^c} = (\frac{a}{b})^c$. In your case, this means $\frac{\sqrt{108 x^{10}}}{\sqrt{2x}} = \sqrt{\frac{108x^{10}}{2x}}$.