Question

What is the following product? RootIndex 5 StartRoot 4 x squared EndRoot times RootIndex 5 StartRoot 4 x squared EndRoot 4 x squared RootIndex 5 StartRoot 16 x Superscript 4 Baseline EndRoot 2 (RootIndex 5 StartRoot 4 x squared EndRoot) 16 x Superscript 4

Answer 1

Answer:

The answer is B on EDGE 2020

Answer 2

Answer:

$(B)\sqrt[5]{16x^4}$

Step-by-step explanation:

We are required to evaluate:

$\sqrt[5]{4x^2} X \sqrt[5]{4x^2}$

By laws of indices: $\sqrt[n]{x}=x^{^\frac{1}{n} }$

Therefore: $\sqrt[5]{4x^2} =(4x^2)^{^\frac{1}{5}$

Thus:

$\sqrt[5]{4x^2} X \sqrt[5]{4x^2}=(4x^2)^{1/5}X(4x^2)^{1/5}\\ Applying same base law of indices:a^mXa^n=a^{m+n}\\(4x^2)^{1/5}X(4x^2)^{1/5}=(4x^2)^{1/5+1/5}=(4x^2)^{2/5}\\ Now, by index product law: a^{mn}=(a^m)^n\\(4x^2)^{2/5}=[(4x^2)^2]^{1/5}=[16x^4]^{1/5}$

$[16x^4]^{1/5}=\sqrt[5]{16x^4} \\ Therefore:\\\sqrt[5]{4x^2} X \sqrt[5]{4x^2}=\sqrt[5]{16x^4}$