## TeemoTheTerific 2 years ago evaluate (a^4)^3 when a = (2r3)^1r6 Ill rewrite it neater

1. TeemoTheTerific

$\huge(a ^{4})^{3}$

2. TeemoTheTerific

$\huge a= (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}$

3. TeemoTheTerific

evaluate

4. TeemoTheTerific

i just sub a equal whatever in a^12 but i can never get it right

5. jim_thompson5910

(a^4)^3 = a^(4*3) (a^4)^3 = a^(12) So if you want to find (a^4)^3, then you can find a^(12) instead since they are the same

6. jim_thompson5910

$\Large a= (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}$ $\Large a^{12}$ $\Large \left((\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\right)^{12}$ $\Large (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 } * 12}$ $\Large (\frac{ 2 }{ 3 })^{\frac{ 12 }{ 6 }}$ $\Large (\frac{ 2 }{ 3 })^{2}$ I'll let you finish

7. TeemoTheTerific

so the final answer is 4r9?

8. TeemoTheTerific

wow thanks so much anyways

9. jim_thompson5910

you mean 4/9 right?

10. jim_thompson5910

not sure how/where the r comes into play

11. jim_thompson5910

btw $\Large \frac{4}{9} = 4/9$

12. TeemoTheTerific

oh lol in my calc to write something over something its something r something :P

13. jim_thompson5910

oh interesting, never seen or heard of that before

14. TeemoTheTerific

actaully some question written in australia are written 8r9 which means 8/9

15. jim_thompson5910

i gotcha, that makes more sense now

16. jim_thompson5910

never ever seen that notation before