Answer:
[tex](B)\sqrt[5]{16x^4}[/tex]
Step-by-step explanation:
We are required to evaluate:
[tex]\sqrt[5]{4x^2} X \sqrt[5]{4x^2}[/tex]
By laws of indices: [tex]\sqrt[n]{x}=x^{^\frac{1}{n} }[/tex]
Therefore: [tex]\sqrt[5]{4x^2} =(4x^2)^{^\frac{1}{5}[/tex]
Thus:
[tex]\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}\\[/tex]
[tex][16x^4]^{1/5}=\sqrt[5]{16x^4} \\$Therefore:\\\sqrt[5]{4x^2} X \sqrt[5]{4x^2}=\sqrt[5]{16x^4}[/tex]
