# very hard question

• Oct 22nd 2009, 08:06 AM
andyboy179
very hard question
hi this is m question:

2X $81^\frac{1}{4}$
• Oct 22nd 2009, 08:08 AM
Quote:

Originally Posted by andyboy179
hi this is m question:

2x $81^\frac{1}{4}$

hint : $3^4=81$
• Oct 22nd 2009, 08:11 AM
andyboy179
dunno, how would i do it?
• Oct 22nd 2009, 08:14 AM
Quote:

Originally Posted by andyboy179
dunno, how would i do it?

$
2\times (3^4)^{\frac{1}{4}}
$
• Oct 22nd 2009, 08:19 AM
andyboy179
i don't understand!!!!!!!!!!!

this is a question which i have done:
3X $125^\frac{1}{3}$ = 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?
• Oct 22nd 2009, 08:29 AM
Quote:

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

this is a question which i have done:
3X $125^\frac{1}{3}$ = 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?

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

so in your case now ,

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

Then proceed , $2\times 3 =6$
• Oct 22nd 2009, 08:33 AM
andyboy179
Quote:

Originally Posted by mathaddict
$5\times 5\times 5=5^3=125$ , therefore , $\sqrt[3]{125}=5$

so in your case now ,

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

Then proceed , $2\times 3 =6$

thankyou very much for all the question you have helped me on(Clapping)(Clapping)
• Oct 22nd 2009, 08:35 AM