Shinichi72
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- Joined
- Apr 2, 2021
- Messages
- 13
So, which means the only error in the problem is problem #2, right?Looking at the image, the error in each is with the final number.
1. As the number outside the radical is 2, within the radical sign it must be [MATH]2^6[/MATH] i.e. 64 (since [MATH]\sqrt[6]{2^6} = 2^\tfrac{6}{6} = 2[/MATH])
2. The number outside the bracket is 2, so within the bracket should be [MATH]2^5[/MATH] i.e. 32 (since [MATH](2^5)^\tfrac{1}{5} = 2^\tfrac{5}{5} = 2[/MATH])
Oh, I see, but can you give me the steps by steps on how did you arrive at your given answer. I really appreciate your effort.There's a problem in both #1 and #2.
1. [MATH]\hspace2ex \sqrt[6]{64x^2y^5}[/MATH]2.[MATH]\hspace2ex \left( 32x^2y^3 \right)^{\tfrac{1}{5}}[/MATH]for the reasons given above.
(Assuming this is what they consider to be simplest form).
Thank you very much1.
[MATH] \begin{align*} \hspace2ex 2\left( x^{\frac{1}{3}} y^{\frac{5}{6}} \right) & =64^{\frac{1}{6}}x^{\frac{1}{3}}y^{\frac{5}{6}}\\ & =64^{\frac{1}{6}}x^{\frac{2}{6}}y^{\frac{5}{6}}\\ & =(64x^2y^5)^{\frac{1}{6}}\\ & =\sqrt[6]{64x^2y^5}\\ \end{align*}[/MATH]
2.
[MATH] \begin{align*} \hspace2ex 2\sqrt[5]{x^2y^3} & =32^\frac{1}{5}x^\frac{2}{5}y^\frac{3}{5}\\ & =(32x^2y^3)^\frac{1}{5}\\ \end{align*}[/MATH]