how do I figure out this problem : 4 ouside sqrt of 81x^4?

Question

The problem is the number four is outside of the sqrt symbol I dont know what do call it but the four is in the same place a 3 would be if I was trying to figure out the cubed root. Then 81x^4 is on the inside of the sqrt symbol. I hope that made it a little easier to understand

Answer 1

We are finding the 4th root:-
(81 x^4)^(1/4) = 3 x

Answer 2

With the four in that place,you know to find the fourth root. This means that your answer can be multiplied by iself four times to get the 81x^4. 81x^4 can also be written as (9*9*x^2*x^2)=(3*3*3*3*x*x*x*x)
Now you have factored the inside out to its lowest prime integers. Now you look at the factorization and see which number or variable occurs four times. A "3" occurs four times, so the fourth root of 81 is three, and "x" occurs four times, so the fourth root of "x^4" is x.
You end up with 3x.

Answer 3

I think you are trying to say the fourth root of 81x^4

Another way to say it would be.. (81x^4)^1/4

That's like saying what is the square root of x squared....here's why...

You have (3^4)(x^4) for 81x^4

Taking the fourth root gives 3x

Answer 4

it is simple surds . cube root of 27 =(3*3*3) is 3
similarly fourth root of 81x^4 =(3*3*3*3*x*x*x*x) is 3x
for more details go for Google search

Answer 5

i think the answer is 3x..