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

b in edge

Explanation:

Answer 2

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

Explanation:

We are required to evaluate:

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

By laws of indices:
`\sqrt[n]{x}=x^{^(1)/(n) }`

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