Mathematics, 20.07.2019 09:30 Ryleetarver
Pls give an explanation w your answer
Answers: 2
![a^{\frac{1}{n}}=\sqrt[n]{a}\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}](/tpl/images/0111/1985/f82b7.png)
![64^\frac{1}{4}=\left(2^6\right)^\frac{1}{4}=2^{6\cdot\frac{1}{4}}\\\\=2^\frac{6}{4}=2^{1\frac{2}{4}}=2^{1+\frac{2}{4}}=2\cdot2^\frac{2}{4}\\\\=2\sqrt[4]{2^2}=2\sqrt[4]4](/tpl/images/0111/1985/8cee6.png)
![64^\frac{1}{4}=\sqrt[4]{64}=\sqrt[4]{16\cdot4}=\sqrt[4]{16}\cdot\sqrt[4]4=2\sqrt[4]4](/tpl/images/0111/1985/846e5.png)
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