how many times can a number be square rooted until it reaches a base (like 1,2, or 3)?

edit: I give fg.

This post was edited by MsRailgun on Sep 20 2013 01:07pm

the answear is 5 if you are going by the principal root rule

Liar.

you didnt specify the type of math you are doing there are many different rules when you are square rooting, cubic - princpal - negitives - based.. please explain the math problem better so i may hlep you

i think he is asking for the digital root

I think I reached an answer - not sure if it's right but keep them answers coming.

if you progressively square root a number and then the result, you will progress towards 1 for any positive number you start with

That is true. But the question remains, how many times can you do it for a particular number until it reaches under, say, 3 or 2 or 1?

This post was edited by MsRailgun on Sep 20 2013 03:22pm

use log to work it out for 3 and 2
ie for a number where it is possible to reach exactly those number or the closest you can get to it
to reach 2 or 3 exactly would need to start with a (2^n)th power of the number
for one it would be infinitely often

This post was edited by brmv on Sep 20 2013 03:33pm

EDIT: PLEASE SEE MY FIXED SOLUTION ON NEXT PAGE

Let's say c is the number we want to reach

and we have number k


if we root k n times, it should be the equivalent of this

k^(1/2n)

we want to know n such that this is < c

k^(1/2n) < c


1/(2n) * log k < c

2n > 1/c * log k

n > 1/(2c) * log k

so you take the smallest n such that n is an integer and is greater than this expression.

This post was edited by zackill4 on Sep 20 2013 03:37pm