My bad I forgot to log c!, so change my answer by replacing c with log c please lol

FIXED:

k^(1/2n)
we want to know n such that this is < c
k^(1/2n) < c
1/(2n) * log k < log c

2n > log k/log c

n > 1/2* log k/log c

smallest integer n such that n > 1/2 * log k / log c

Reminder: n is the NUMBER of square roots it takes. k is the number we're square rooting and c is the "base". Checked it with 4 and 2, seems legit :/

Note for c = 1 that this expression is infinite (because no such n exists, unless k = 1, but then the expression is undefined)

This post was edited by zackill4 on Sep 20 2013 03:51pm

2^n is not correct because we are dealing with square rooting. It would have to be 2^(2n)...

For example, 8 = 2^3, yet you can't square root 8 an integer number of times to get exactly 2


Basically, this problem boils down to a simple log, except that instead of looking for the exponent x such that a^x = b, we want to restrict x to be multiple of 2
The problem then becomes a^(2x) > b, which is why the solution is > half of log base a of b

This post was edited by zackill4 on Sep 20 2013 04:00pm

guess you misunderstood what i was saying
"start with a (2^n)th power of the number" means x^(2^n)

2^(2n) is not correct, eg 2^6 = 128 and it's square root is 11.3137....

lol, my bad, idk what I was thinking


FIXED AGAIN:

k^(1/2^n)
we want to know n such that this is < c
k^(1/2^n) < c
1/(2^n) * log k < log c

2^n > log k/log c
log both sides

n > 1/log(2)* log(log k/log c)

This post was edited by zackill4 on Sep 20 2013 04:45pm