Mathematics, 05.05.2020 08:25 loudenalexisp56lp0
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
Answers: 2
![(B)\sqrt[5]{16x^4}](/tpl/images/0635/3839/7387a.png)
![\sqrt[5]{4x^2} X \sqrt[5]{4x^2}](/tpl/images/0635/3839/b0078.png)
![\sqrt[n]{x}=x^{^\frac{1}{n} }](/tpl/images/0635/3839/7238f.png)
![\sqrt[5]{4x^2} =(4x^2)^{^\frac{1}{5}](/tpl/images/0635/3839/08c5c.png)
![\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}\\](/tpl/images/0635/3839/ddae2.png)
![[16x^4]^{1/5}=\sqrt[5]{16x^4} \\$Therefore:\\\sqrt[5]{4x^2} X \sqrt[5]{4x^2}=\sqrt[5]{16x^4}](/tpl/images/0635/3839/9286a.png)
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What is the following product? RootIndex 5 StartRoot 4 x squared EndRoot times RootIndex 5 StartRoot...
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