Question

lets do some more problems with higher order roots
problem 3.1
(sqrt7{128}=) (\boxed{quad})
check
explain

Explanation:

Step1: Factor 128

We know that \(128 = 64\times2\), and \(64 = 2^6\). So we can rewrite \(128\) as \(2^6\times2=2^7\).

Step2: Simplify the seventh - root

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (for \(a\geq0,b\geq0\)) and \(\sqrt[n]{x^m}=x^{\frac{m}{n}}\) (when \(x\geq0\)), we have \(\sqrt[7]{128}=\sqrt[7]{2^7}\).
According to the property \(\sqrt[n]{x^n}=x\) (for \(x\geq0\) and \(n\) is a positive integer), when \(n = 7\) and \(x = 2\), \(\sqrt[7]{2^7}=2\).

Answer:

\(2\)