## yugobister Group Title 27^2/3 2 years ago 2 years ago

1. ns36 Group Title

I'm assuming you mean 27^(2/3). If so, you first take the cube root of 27 (solve for the number which, cubed, yields 27). This number is 3. Then you square 3 to get 9. 9 is therefore the answer. Make sense?

2. yugobister Group Title

kind of so if i do my next question: 64^(5/6). Then im not completely sure on what the next step would be.

3. ns36 Group Title

The answer is 32. First, you have to find the number x that when taken to the sixth power yields 64. So x^6 should be 64.

4. Rogenhamen Group Title

$27^{2/3}=\sqrt[3]{27^2}$$\sqrt[3]{27^2}=\sqrt[3]{729}$$\sqrt[3]{729}=9$

5. ns36 Group Title

In this case, that number is 2. Next you should take that x and take it to the fifth power (x^5) to find your final answer, which is 32.

6. Rogenhamen Group Title

$64^{5/6}=\sqrt[5]{64^6}$$\sqrt[6]{64^5}=\sqrt[6]{1073741824}$$\sqrt[6]{1073741824}=32$

7. yugobister Group Title

i am still kind of confused. so for 1^5/3.

8. ns36 Group Title

Again, you should first find out what number, when cubed, yields 1. That number here is simply 1. Then you should take that number (in this case, 1) and take it to the 5th power (1^5). So the answer is 1.

9. Rogenhamen Group Title

$1^{5/3}=\sqrt[3]{1^5}$$\sqrt[3]{1^5}=\sqrt[3]{1}$$\sqrt[3]{1}=1$