How to solve? | Nice Math Olympiad Question

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  • Опубліковано 27 січ 2025

КОМЕНТАРІ • 16

  • @noski33
    @noski33 15 днів тому +3

    Something is wrong! a^(m+n) is not equal to a^(m•n); it is equal to a^m•a^n. The steps he wrote after are correct though

    • @MathBeast.channel-l9i
      @MathBeast.channel-l9i  15 днів тому +2

      You are right....that is a typo.
      I realized it now that was written mistakenly and I'm sorry for that 😔

    • @angelbest19lop78
      @angelbest19lop78 15 днів тому

      Is there a way to record video ?? Poor students will get headaches and start hating math because of mistakes or typos mentioned..Please! Amen​@@MathBeast.channel-l9i

    • @angelbest19lop78
      @angelbest19lop78 15 днів тому

      ​@@MathBeast.channel-l9isorry re-record and delete video I meant

  • @srinivasch-re2oq
    @srinivasch-re2oq 14 днів тому +1

    Very simple, 6^x (36 -1) = 60
    6^x = 60/35 = 12/7
    Apply logerthms then find x value.
    xlog6 = log(12/7)
    We should know logerthm, log inverse values from 1 to 20 while preparing for
    such exams.

  • @alfredhorstman
    @alfredhorstman 15 днів тому +1

    O common!!! That can be much shorter!!!

  • @에스피-z2g
    @에스피-z2g 15 днів тому

    6^x(36-1)=60
    6^x=12/7
    x=(log12-log7)/log6
    x=1+(log2-log7)/log6
    x=1-(log7-log2)/log6

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 16 днів тому +1

    6^{x+x ➖ }+{2+2 ➖} ➖ 36 =6^{x^2+4} ➖ 36=6^2x^2 ➖ 36 36x^2 ➖ 36={1.24+1.124}=2.48 2.24^24 2.2^12^2^12 1.1^6^6 2^3^2^3 2^1^1^3 23 (x ➖ 3x+2).

  • @Skywarddecrypter
    @Skywarddecrypter 15 днів тому +1

    Why u increased unnecessary steps?

  • @MrThetaphi
    @MrThetaphi 15 днів тому +1

    At 0:29 you write a^(m+n) = a^(m*n) which feels terribly wrong, also at 7:40. Luckily, you didn't use this, else you'd have gotten doomed. Am I totally wrong?

    • @MathBeast.channel-l9i
      @MathBeast.channel-l9i  15 днів тому

      No.
      You are totally right Sir.
      It's a typo.
      Very sorry for that it was mistaken 😔
      It's actually a^(m+n) = a^m × a^n