# Solve [(86400*n^2)/2^n]<1 solve n

Asked Jan 29, 2006, 08:18 AM — 3 Answers
Could you Please solve this problem

[(86400*n^2)/2^n]<1 solve n

 eawoodall Posts: 232, Reputation: 48 Full Member #2 Jan 31, 2006, 04:24 AM
Notice that
2^1 =2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
2^10>1k
2^11 >2k
2^12>4k
2^13>8k
2^14>16k
2^15>32k
2^16>64k
2^17>128k
2^18>256k
2^19>512k
2^20>1M
2^21>2M
2^22=4,194304>4M

Also notice
1^2=1
2^2=4
3^2=9
15^2=225
17^2=289
18^2=324
21^2=441
22^2=484

22^2 * 86400 = 484 * 86400 = 4,1817,600.

86400 * n^2 < 2^n when 86.k * n^2 < 2^n.

So at n=22 2^n > n^2 * 86400.

I hope that helps.
 Rehaan_genius Posts: 49, Reputation: 4 Junior Member #3 Aug 11, 2008, 01:38 AM
In left hand side of the inequality, the denominator is always positive for all values of n so we can cross multiply, which gives:-
86400*n^2<2^n. So then find that value of n for which the inequality holds. I think you should first try yourself now!
 saw2005 Posts: 1, Reputation: 1 New Member #4 Oct 5, 2012, 06:47 PM
whats 10% of 300

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