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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. :(

Thanks in advance guys. Seriously,

JD. :)

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    $\begingroup$ I do not see any "equation" to "solve." Do you mean you want to simplify the expression? $\endgroup$ – Rory Daulton Aug 23 '15 at 16:20
  • $\begingroup$ Please learn about indices and exponentiation. $\endgroup$ – Narasimham Aug 23 '15 at 16:26
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I believe the simplest way to simplify your expression is with fractional exponents, though there are other ways.

$$\begin{align} \sqrt[10]{32a^5} &= \left(32a^5\right)^{1/10} \\[2 ex] &= \left(2^5a^5\right)^{1/10} \\[2 ex] &= 2^{5\cdot 1/10}a^{5\cdot 1/10} \\[2 ex] &= 2^{1/2}a^{1/2} \\[2 ex] &= (2a)^{1/2} \\[2 ex] &= \sqrt{2a} \end{align}$$

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  • $\begingroup$ Oh lordy it was that simple hey? Thanks so much !!!!! you guys are the best. $\endgroup$ – Jane DeRulo Aug 24 '15 at 0:54
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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 $$ \sqrt[10]{32a^5}=\sqrt{\sqrt[5]{(2a)^5}}=\sqrt{2a} $$

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