Question

1. simplify \\(\sqrt4{81x^{17}}\\) completely given \\(x > 0\\).

Explanation:

Step1: Factor the radicand

We can factor \(81\) and \(x^{17}\) to find perfect fourth - power factors.
We know that \(81 = 3^4\) and \(x^{17}=x^{4\times4 + 1}=x^{16}\cdot x\) (since when we divide the exponent of \(x\) by \(4\), \(17\div4 = 4\) with a remainder of \(1\), so \(x^{17}=(x^{4})^4\cdot x\)).
So, \(\sqrt[4]{81x^{17}}=\sqrt[4]{3^{4}\cdot x^{16}\cdot x}\)

Step2: Use the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\))

According to the property \(\sqrt[4]{3^{4}\cdot x^{16}\cdot x}=\sqrt[4]{3^{4}}\cdot\sqrt[4]{x^{16}}\cdot\sqrt[4]{x}\)

Step3: Simplify each radical

We know that \(\sqrt[4]{3^{4}} = 3\) (because \(\sqrt[n]{a^{n}}=a\) when \(n\) is even and \(a\geq0\)), and \(\sqrt[4]{x^{16}}=\sqrt[4]{(x^{4})^{4}}=x^{4}\) (using the property \(\sqrt[n]{a^{m}}=a^{\frac{m}{n}}\), here \(n = 4\), \(m = 16\), so \(x^{\frac{16}{4}}=x^{4}\)). And \(\sqrt[4]{x}=x^{\frac{1}{4}}\)

So, \(\sqrt[4]{3^{4}}\cdot\sqrt[4]{x^{16}}\cdot\sqrt[4]{x}=3\cdot x^{4}\cdot\sqrt[4]{x}=3x^{4}\sqrt[4]{x}\)

Answer:

\(3x^{4}\sqrt[4]{x}\)