Japan | A Nice Algebra Problem | Math Olympiad

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  • čas přidán 29. 08. 2024
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Komentáře • 13

  • @cybisz2883
    @cybisz2883 Před měsícem

    1:06 It's actually easier to solve if you take a second log at this point. If we have 4^x ⋅ log9 = 9^x ⋅ log(4), then we can take a second log and write log(4^x ⋅ log9) = log(9^x ⋅ log4). Using the law that log(ab) = loga + logb and the law that log(a^x) = x⋅log(a), we can write this as xlog4 + log9 = xlog9 + log4. Now its just solving a linear equation of one variable. Combined with the rule that log(a/b) = loga - logb (just for simplifying, not really necessary for solving), this gives us that x = log(log4 / log9) / log(4/9), or x ≈ 0.5679412839581517.

  • @SidneiMV
    @SidneiMV Před měsícem

    4ˣln9 = 9ˣln4
    2²ˣln3 = 3²ˣln2
    (3/2)²ˣ = (1/ln2)ln3
    2x[ln(3/2)] = ln[(1/ln2)ln3]
    *x = (1/2)[ln( ln3 - ln2)]/(ln3 - ln2)*

  • @friskel9459
    @friskel9459 Před měsícem +1

    Or you do => 9^(4^x)^x = 4^(9^x)^x => (9^x)^(4^x) = (4^x)^(9^x) => (9^x)^(1/9^x) = (4^x)^(1/4^x) => 4^x = 9^x => x = 0

    • @cybisz2883
      @cybisz2883 Před měsícem

      x = 0 would imply that 4 = 9. That first implication of yours, giving (9^x)^(4^x) = (4^x)^(9^x) I'm pretty sure is mistaken.

    • @friskel9459
      @friskel9459 Před měsícem

      @@cybisz2883every number n at n^0 equal 1 so my answer isn't wrong. And my first implication is based on (n^a)^b = n^ab = (n^b)^a

    • @cybisz2883
      @cybisz2883 Před měsícem

      @@friskel9459 9^4^x is supposed to be interpreted as 9^(4^x). You've accidentally interpreted it as (9^4)^x which is not the same thing.

    • @friskel9459
      @friskel9459 Před měsícem

      @@cybisz2883 where did I do that ?

    • @cybisz2883
      @cybisz2883 Před měsícem

      @@friskel9459 In your assertion just now that x=0 is a valid solution, using the property that (n^a)^b = n^(ab). It *is* a valid solution if the problem is interpreted as (9^4)^x = (4^9)^x. However, it is not a valid solution if the problem is interpreted as 9^(4^x) = 4^(9^x) as intended. Why? 9^(4^0) = 9^1 = 9 and 4^(9^0) = 4^1 = 4. Thus, 9^(4^x) = 4^(9^x) would simplify to 9 = 4 if we plug in x=0.

  • @Nguyễn-j9q
    @Nguyễn-j9q Před měsícem +1

    Well just count it
    9⁴=6561
    4⁹=262144
    and to made
    These number equal to each other
    X will be={0}

    • @doichecabano
      @doichecabano Před měsícem +3

      Башня степеней считается сверху вниз, а не снизу вверх.

    • @cybisz2883
      @cybisz2883 Před měsícem

      9 ^ 4 ^ x is supposed to be interpreted as 9 ^ (4 ^ x). You've accidentally assumed it was equal to (9 ^ 4) ^ x which is not the same thing. Note that if you plug your solution, x = 0, into the original equation and simplify, we get that 4 = 9 which is obviously false. Thus 0 cannot be a solution.