how to compare exponents having large powers

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  • Опубліковано 14 жов 2024

КОМЕНТАРІ • 12

  • @rommelsingh1104
    @rommelsingh1104 Рік тому

    Thanks for great presentation. I learned your math technique. THANK YOU!!

  • @adaking8092
    @adaking8092 2 роки тому +1

    much superb. wow.👍👍👍

  • @raghavanaiduvelakaturi5523
    @raghavanaiduvelakaturi5523 3 роки тому +2

    wow sir it helped me a
    lot thanks

  • @mant4309
    @mant4309 2 роки тому +1

    Nice explanation. Although, solution to the second problem is incorrect. The correct solution, i think, is as follows:
    2^10 = 1024 > 10^3
    Taking 5th power on both sides, (2^10)^5 > (10^3)^5
    => 2^50 > 10^15
    => n = 50.
    Although, it is still not enough to say that 50 is the LEAST integer to satisfy the ineqality.

    • @AmanPhogat.
      @AmanPhogat. Рік тому

      Ya if instead of 15 there is 300 than
      By this we end up to 2 ^ 1000 but 997 can satisfy this inequality.

  • @Alexander-oh6ps
    @Alexander-oh6ps 2 роки тому

    Gold video

  • @cegprakash
    @cegprakash 3 роки тому +1

    Great explanation for the first comparison problem. But for the second problem
    2^50 > 10^15
    2^(50/15) is also > 10..
    So shouldn't n be 50 sir?

    • @LarsMartinGihle
      @LarsMartinGihle 2 роки тому

      He said that n should be an integer, but i think he meant that n/15 has to be an integer

  • @saiharsha5945
    @saiharsha5945 2 роки тому

    Thank you very much

  • @AmanPhogat.
    @AmanPhogat. Рік тому +1

    No sir 9:50
    Least integral value will be 50 👍

  • @lokeshudayagiri3893
    @lokeshudayagiri3893 4 роки тому +1

    superb sir..

  • @stran006
    @stran006 3 роки тому

    thank you, merci, gracias