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I have an expression I am to simplify:

$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$

I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:

$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ = $\frac{15*\sqrt[4]{5}*\sqrt[4]{25}}{\sqrt[4]{5}}$ =

(cancel out $\sqrt[4]{5}$ present in both numerator and denominator) leaving: $$15\sqrt[4]{25}$$

Where did I go wrong and how can I arrive at $15\sqrt{5}$?

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    $\begingroup$ The fourth root is the square root of the square root ... $\endgroup$ – Ethan Bolker Jan 3 '19 at 16:54
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$$\frac{\sqrt[4] {125}}{\sqrt[4] 5} = \frac{\sqrt[4] {5^3}}{\sqrt[4] 5} = \sqrt[4]{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$

Even quicker, $\sqrt[4]{25}$ means $\sqrt{\sqrt{25}}$, which becomes $\sqrt{5}$.

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Write $$\sqrt[4]{\frac{125}{5}}=\sqrt[4]{25}=\sqrt{5}$$

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It turns out that $\sqrt[4]{25}=\sqrt{5}$. This is because $25=5^2$, so that $\sqrt[4]{25}=\sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$ So, you are correct, as is the book.

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  • $\begingroup$ Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me. $\endgroup$ – Doug Fir Jan 3 '19 at 17:09

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