I have no idea how to do this. Any ideas anyone
2007-05-23 13:47:01 · 5 answers · asked by got.pocky 2 in Science & Mathematics ➔ Mathematics
( 2 / 2)^2 + N - 1
2007-05-23 13:51:28 · answer #1 · answered by Renaissance Man 5 · 2⤊ 2⤋
0 = (2-2)*2
1 = 2^(2-2)
2 = 2*2/2
3 = 2+ 2/2
6 = 2+2+2
8 = 2*2*2
There are many many more
2007-05-23 21:06:44 · answer #2 · answered by Ohil 3 · 0⤊ 1⤋
log[ log(2) / log( âââ...â2) ] / log(2) = N where N is the number of square root signs. [for N=0 use 2*(2-2)]
Proof:
â2 = 2^(1/2)
ââ2 = (2^(1/2))^(1/2) = 2^(1/4)
âââ2 = (2^(1/4))^(1/2) = 2^(1/8)
...
âââ...â2 = 2^(1/(2^N)) where N is the number of square roots
take the log_2 (log base 2) of both sides
log_2(âââ...â2) = 1/(2^N)
log(âââ...â2) / log(2) = 1/(2^N)
log(2) / log(âââ...â2) = 2^N
log_2 [log(2) / log(âââ...â2)] = N
log[log(2) / log(âââ...â2)] / log(2) = N
2007-05-23 21:10:47 · answer #3 · answered by Scott R 6 · 1⤊ 0⤋
sqrt(2/2) x N
The sqrt symbol can have that little 2 just above it.
And 1 x N = N
Or, yes (2/2)^2 x N.
2007-05-23 21:02:39 · answer #4 · answered by BotanyDave 5 · 0⤊ 2⤋
yes. and as you asked i wrote it down. now what do i do?
2007-05-23 20:49:58 · answer #5 · answered by Anonymous · 0⤊ 3⤋