# Simplify $\,\sqrt[10]{32a^5}$

I'm not sure if this is the correct site to ask such an elementary question but I'm trying to teach myself basic algebra and I can't understand how to do this one equation it's been so annoying.

So basically this is the expression:

$$\sqrt[10]{32a^5}$$

What I thought you're meant to do is simply it to $\sqrt{32a}$ then further to $4\sqrt{2a}$. But, I know the answer is $\sqrt{2a}$.

Lord help me. Perhaps I'm just not meant to do this kind of stuff. :(

JD. :)

• I do not see any "equation" to "solve." Do you mean you want to simplify the expression? – Rory Daulton Aug 23 '15 at 16:20
• Please learn about indices and exponentiation. – Narasimham Aug 23 '15 at 16:26

\begin{align} \sqrt[10]{32a^5} &= \left(32a^5\right)^{1/10} \\[2 ex] &= \left(2^5a^5\right)^{1/10} \\[2 ex] &= 2^{5\cdot 1/10}a^{5\cdot 1/10} \\[2 ex] &= 2^{1/2}a^{1/2} \\[2 ex] &= (2a)^{1/2} \\[2 ex] &= \sqrt{2a} \end{align}
Since $(x^2)^5=x^{10}$, we can deduce that $$\sqrt[10]{y}=\sqrt{\sqrt[5]{y}}$$ because the tenth power of both sides is the same (for $x>0$ and $y>0$): $$(\sqrt[10]{y})^{10}=y$$ and $$(\sqrt{\sqrt[5]{y}})^{10}=((\sqrt{\sqrt[5]{y}})^2)^5= (\sqrt[5]{y})^5=y$$
Since $32=2^5$, we can write $32a^5=2^5a^5=(2a)^5$, so $$\sqrt[10]{32a^5}=\sqrt{\sqrt[5]{(2a)^5}}=\sqrt{2a}$$