## anonymous 4 years ago Simplify the expression and eliminate any negative exponents, assuming that all letters denote positive numbers. (2x^4 * y^(-4/5))^3(8y^2)^(2/3) I try to post it in equation form too.

1. anonymous

Could you draw it out?

2. anonymous

$(2x^4y^\frac{ -4 }{ 5 })^3(8y^2)^\frac{ 2 }{ 3 }$

3. anonymous

if it was $(a ^{4})^{3}$ What have you learned to do with the exponents?

4. anonymous

Multiply 4 and 3, right?

5. anonymous

Yes, what if it was $(4a ^{4})^{3}$ ?

6. anonymous

4 to the third times a to the 12

7. anonymous

Yes, what happens with $(8y ^{2})^{2/3}$ ?

8. anonymous

$8^\frac{ 2 }{ 3 }y ^\frac{ 4 }{ 3 }$

9. anonymous

Ok, so which part do you have trouble with?

10. anonymous

Well, the answer in the back of the book is 14, and no matter what I do, i can't get 14! I'll take a picture of my work real quick and maybe you can see what I did then?

11. anonymous

Sure

12. helder_edwin

$\large (2x^4y^{-4/5})^3(8y^2)^{2/3}=2^3x^{4\times3}y^{-4/5\times3} 8^{2/3}y^{2\times2/3}$

13. helder_edwin

$\large 8x^{12}y^{-12/5}8^{2/3}y^{4/3}$

14. helder_edwin

$\large =8^{1+2/3}x^{12}y^{4/3-12/5}=8^{5/3}x^{12}y^{-16/15}$

15. lopus

|dw:1347151652056:dw|

16. lopus

$2x^{12}*\frac{ 8 }{ y}$

17. anonymous

Here's what I have... I've just tried a few times and I'm really having difficulty. This is my most recent bit.

18. helder_edwin

$\large =2^5x^{12}y^{-16/15}$

19. anonymous

According to the back of the book, the answer is $(32x^\frac{ 12 }{ 1 })/(y^ \frac{ 16 }{ 15 })$

20. anonymous

I just don't have an idea of how to get it.

21. helder_edwin

$\large \frac{32x^{12}}{y^{16/15}}$ 2^5=32

22. lopus

|dw:1347152276782:dw|

23. anonymous

You broke down your y's by saying $y ^{4/3}= y \sqrt[3]{y}$ and such...

24. lopus

$y ^{-1}=\frac{ 1 }{ y }$

25. anonymous

|dw:1347152573807:dw|

26. anonymous

ok, thanks everyone! i have a better understanding now.