Question

$sqrt4{x^{3}} cdot sqrt{x}$

Explanation:

Step1: Convert radicals to exponents

Recall that \(\sqrt[n]{a}=a^{\frac{1}{n}}\). So, \(\sqrt[4]{x^{3}}=(x^{3})^{\frac{1}{4}}\) and \(\sqrt{x}=x^{\frac{1}{2}}\).

Step2: Simplify the exponents

Using the power - of - a - power rule \((a^{m})^{n}=a^{mn}\), we have \((x^{3})^{\frac{1}{4}} = x^{\frac{3}{4}}\) and \(x^{\frac{1}{2}}\) remains as it is. Now, when we multiply two exponential expressions with the same base \(a^{m}\cdot a^{n}=a^{m + n}\), so \(x^{\frac{3}{4}}\cdot x^{\frac{1}{2}}=x^{\frac{3}{4}+\frac{1}{2}}\).

Step3: Add the exponents

First, find a common denominator for \(\frac{3}{4}\) and \(\frac{1}{2}\). The common denominator of 4 and 2 is 4. So, \(\frac{1}{2}=\frac{2}{4}\). Then \(\frac{3}{4}+\frac{2}{4}=\frac{3 + 2}{4}=\frac{5}{4}\). So, \(x^{\frac{3}{4}+\frac{1}{2}}=x^{\frac{5}{4}}\).

Step4: Convert back to radical form (optional, but to match the original form)

Since \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\), \(x^{\frac{5}{4}}=\sqrt[4]{x^{5}}\). We can also simplify \(\sqrt[4]{x^{5}}\) as \(\sqrt[4]{x^{4}\cdot x}\). Using the property \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\)), we get \(\sqrt[4]{x^{4}\cdot x}=\sqrt[4]{x^{4}}\cdot\sqrt[4]{x}=x\sqrt[4]{x}\)

Answer:

\(x\sqrt[4]{x}\) (or \(\sqrt[4]{x^{5}}\))