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-   -   Solve [(86400*n^2)/2^n]<1 solve n (https://www.askmehelpdesk.com/showthread.php?t=19084)

  • Jan 29, 2006, 09:18 AM
    kruthi_om
    solve [(86400*n^2)/2^n]<1 solve n
    Could you Please solve this problem

    [(86400*n^2)/2^n]<1 solve n
  • Jan 31, 2006, 05:24 AM
    eawoodall
    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.
  • Aug 11, 2008, 01:38 AM
    Rehaan_genius
    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!
  • Oct 5, 2012, 06:47 PM
    saw2005
    What's 10% of 300

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