Question

question
simplify (sqrt4{1296x^{40}}) completely given (x > 0).
answer attempt 1 out of 2

Explanation:

Step1: Simplify the constant term

We know that \(1296 = 6^4\) since \(6\times6\times6\times6 = 1296\). So we can rewrite the fourth - root of \(1296\) as \(\sqrt[4]{6^4}\). By the property of \(n\) - th roots, \(\sqrt[n]{a^n}=a\) when \(n\) is even and \(a\geq0\). Here \(n = 4\) (even) and \(6>0\), so \(\sqrt[4]{6^4}=6\).

Step2: Simplify the variable term

For the term with \(x\), we have \(\sqrt[4]{x^{40}}\). Using the property of exponents and roots \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), here \(n = 4\) and \(m = 40\). So \(x^{\frac{40}{4}}=x^{10}\). We can also think about it in terms of the \(n\) - th root of a power: \(\sqrt[4]{x^{40}}=\sqrt[4]{(x^{10})^4}\), and since \(x>0\), by the property \(\sqrt[n]{a^n}=a\) (for even \(n\) and \(a > 0\)), we get \(\sqrt[4]{(x^{10})^4}=x^{10}\).

Step3: Combine the results

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}\) (for \(a\geq0,b\geq0\) and \(n\) a positive integer), we have \(\sqrt[4]{1296x^{40}}=\sqrt[4]{1296}\times\sqrt[4]{x^{40}}\). Substituting the results from Step 1 and Step 2, we get \(6\times x^{10}=6x^{10}\).

Answer:

\(6x^{10}\)