very hard question

**andyboy179** · October 22nd 2009, 08:06 AM

hi this is m question:

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

**mathaddict** · October 22nd 2009, 08:08 AM

Originally Posted by andyboy179

hi this is m question:

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

hint : $3^4=81$

**andyboy179** · October 22nd 2009, 08:11 AM

dunno, how would i do it?

**mathaddict** · October 22nd 2009, 08:14 AM

Originally Posted by andyboy179

dunno, how would i do it?

$<br /> 2\times (3^4)^{\frac{1}{4}}<br />$

**andyboy179** · October 22nd 2009, 08:19 AM

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?

**mathaddict** · October 22nd 2009, 08:29 AM

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 , $2\times 3 =6$

**andyboy179** · October 22nd 2009, 08:33 AM

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

**mathaddict** · October 22nd 2009, 08:35 AM

Originally Posted by andyboy179

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

you are welcome .

Thread: very hard question

LinkBack

Thread Tools

Display

very hard question

Similar Math Help Forum Discussions

cos question hard

Hard question.

Please Help, Hard Question for me.

Need some help with a hard question

need help hard question

Search Tags