1. ## very hard question

hi this is m question:

2X$\displaystyle 81^\frac{1}{4}$

2. Originally Posted by andyboy179
hi this is m question:

2x$\displaystyle 81^\frac{1}{4}$
hint : $\displaystyle 3^4=81$

3. dunno, how would i do it?

4. Originally Posted by andyboy179
dunno, how would i do it?
$\displaystyle 2\times (3^4)^{\frac{1}{4}}$

5. i don't understand!!!!!!!!!!!

this is a question which i have done:
3X$\displaystyle 125^\frac{1}{3}$ = cube root 125= 5x3=15

i need to set it out like this for the question i posted, do you know how to work it out like this?

6. Originally Posted by andyboy179
i don't understand!!!!!!!!!!!

this is a question which i have done:
3X$\displaystyle 125^\frac{1}{3}$ = cube root 125= 5x3=15

i need to set it out like this for the question i posted, do you know how to work it out like this?
$\displaystyle 5\times 5\times 5=5^3=125$ , therefore , $\displaystyle \sqrt[3]{125}=5$

so in your case now ,

$\displaystyle 3\times 3\times 3\times 3=3^4=81$, therefore , $\displaystyle \sqrt[4]{81}=3$

Then proceed , $\displaystyle 2\times 3 =6$

$\displaystyle 5\times 5\times 5=5^3=125$ , therefore , $\displaystyle \sqrt[3]{125}=5$

so in your case now ,

$\displaystyle 3\times 3\times 3\times 3=3^4=81$, therefore , $\displaystyle \sqrt[4]{81}=3$

Then proceed , $\displaystyle 2\times 3 =6$

thankyou very much for all the question you have helped me on

8. Originally Posted by andyboy179
thankyou very much for all the question you have helped me on
you are welcome .