Indian l can you solve this?? l Nice Olympiad Math Exponential Problem

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  • Опубліковано 24 січ 2025

КОМЕНТАРІ • 10

  • @ST-dr6qx
    @ST-dr6qx 15 днів тому

    Great!!! Thanks!!!!

  • @princejag
    @princejag 15 днів тому

    mesmerizing.

  • @TitovaNadejda
    @TitovaNadejda 11 днів тому

    ❤😮😊

  • @robertovaladao3136
    @robertovaladao3136 15 днів тому

    Very good!

  • @mikewinny3856
    @mikewinny3856 11 днів тому +1

    Thank you, Cap!

  • @BrijLalSahu-bb1lu
    @BrijLalSahu-bb1lu 15 днів тому

    Good @MATHS TUTORIAL (BL SAHU)

  • @Марія-75
    @Марія-75 15 днів тому +1

  • @dahupaylo
    @dahupaylo 15 днів тому +1

    Method is unnecessarily complicated.
    4^2a=48
    Take ln both sides:
    2a*ln4=ln(4*4*3)
    2a*ln4=2ln4+ln3
    Divide both sides by 2ln4
    a=1+ln3)/2ln4
    But ln4=2ln2
    a=1+ln3/4ln2

  • @stpat7614
    @stpat7614 15 днів тому

    Including complex roots:
    4^a * 4^a = 48
    4^(a + a) = (48^[1 / 2])^2
    4^(2 * a) = ([2^4 * 3]^[1 / 2])^2
    4^(a * 2) = ([2^4]^[1 / 2] * 3^[1 / 2])^2
    (4^a)^2 = (2^[4 * (1 / 2)] * 3^[1 / 2])^2
    (4^a)^2 = (2^2 * 3^[1 / 2])^2
    (4^a)^2 = (4 * 3^[1 / 2])^2
    sqrt([4^a]^2) = +/- sqrt([4 * 3^(1 / 2)]^2)
    4^a = +/- 4 * 3^(1 / 2)
    (1 / 4) * 4^a = +/- (1 / 4) * 4 * 3^(1 / 2)
    4^a * 4^(-1) = +/- 4^1 * 4^(-1) * 3^(1 / 2)
    4^(a - 1) = +/- 4^(1 - 1) * 3^(1 / 2)
    4^(a - 1) = +/- 4^0 * 3^(1 / 2)
    4^(a - 1) = +/- 1 * 3^(1 / 2)
    4^(a - 1) = +/- 3^(1 / 2)
    (2^2)^(a - 1) = +/- ([3^(1 / 2)]^[1 / 2])^2
    2^(2 * [a - 1]) = +/- (3^[(1 / 2) * (1 / 2)])^2
    2^([a - 1] * 2) = +/- (3^[1 / 4])^2
    (2^[a - 1])^2 = +/- (3^[1 / 4])^2
    (2^[a - 1])^2 = +(3^[1 / 4])^2, or (2^[a - 1])^2 = -(3^[1 / 4])^2
    Let x = 2^(a - 1), and y = 3^(1 / 4)
    (2^[a - 1])^2 = (3^[1 / 4])^2, or (2^[a - 1])^2 = -(3^[1 / 4])^2
    => x^2 = y^2, or x^2 = -y^2
    => x^2 - y^2 = y^2 - y^2, or x^2 + y^2 = -y^2 + y^2
    => x^2 - y^2 = 0, or x^2 + y^2 = 0
    => (x - y)(x + y) = 0, or (x - i * y)(x + i * y) = 0
    => (2^[a - 1] - 3^[1 / 4])(2^[a - 1] + 3^[1 / 4]) = 0, or (2^[a - 1] - i * 3^[1 / 4])(2^[a - 1] + i * 3^[1 / 4]) = 0
    Suppose 2^(a - 1) - 3^(1 / 4) = 0
    2^(a - 1) - 3^(1 / 4) = 0
    2^(a - 1) - 3^(1 / 4) + 3^(1 / 4) = 0 + 3^(1 / 4)
    2^(a - 1) = 3^(1 / 4)
    ln(2^[a - 1]) = ln(3^[1 / 4])
    (a - 1) * ln(2) = (1 / 4) * ln(3)
    (a - 1) * ln(2) / ln(2) = ln(3) / (4 * ln[2])
    (a - 1) * log_2(2) = log_2(3) / 4
    (a - 1) * 1 = log_2(3) / 4
    a - 1 + 1 = log_2(3) / 4 + 1
    a = log_2(3) / 4 + 1
    Suppose 2^(a - 1) + 3^(1 / 4) = 0
    2^(a - 1) + 3^(1 / 4) - 3^(1 / 4) = 0 - 3^(1 / 4)
    2^(a - 1) = -3^(1 / 4)
    ln(2^[a - 1]) = ln(-1 * 3^[1 / 4])
    ln(2^[a - 1]) = ln(i^2 * 3^[1 / 4])
    (a - 1) * ln(2) = ln(i^2 * 3^[1 / 4])
    (a - 1) * ln(2) / ln(2) = ln(i^2 * 3^[1 / 4]) / ln(2)
    (a - 1) * log_2(2) = ln(i^2) / ln(2) + ln(3^[1 / 4]) / ln(2)
    (a - 1) * 1 = ln(e^[i * tau / 2]) / ln(2) + log_2(3^[1 / 4])
    a - 1 = (i * tau / 2) * ln(e) / ln(2) + log_2(3^[1 / 4])
    a - 1 = (i * tau / 2) * 1 / ln(2) + log_2(3^[1 / 4])
    a - 1 = i * tau / (2 * ln[2]) + (1 / 4) * log_2(3)
    a - 1 + 1 = i * tau / (2 * ln[2]) + log_2(3) / 4 + 1
    a = i * tau / (2 * ln[2]) + log_2(3) / 4 + 1
    a = i * 2 * tau / (2^2 * ln[2]) + log_2(3) / 4 + 1
    a = i * 2 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    Suppose 2^(a - 1) - i * 3^(1 / 4) = 0
    2^(a - 1) - i * 3^(1 / 4) = 0
    2^(a - 1) - i * 3^(1 / 4) + i * 3^(1 / 4) = 0 + i * 3^(1 / 4)
    2^(a - 1) = i * 3^(1 / 4)
    ln(2^[a - 1]) = ln(i * 3^[1 / 4])
    (a - 1) * ln(2) = ln(i * 3^[1 / 4])
    (a - 1) * ln(2) / ln(2) = ln(i * 3^[1 / 4]) / ln(2)
    (a - 1) * log_2(2) = ln(i) / ln(2) + ln(3^[1 / 4]) / ln(2)
    (a - 1) * 1 = ln(e^[i * tau / 4]) / ln(2) + (1 / 4) * ln(3^[1 / 4]) / ln(2)
    a - 1 = (i * tau / 4) * ln(e) / ln(2) + ln(3) / (4 * ln[2])
    a - 1 = (i * tau / 4) * 1 / ln(2) + ln(3) / (4 * ln[2])
    a - 1 = i * tau / (4 * ln[2]) + ln(3) / (4 * ln[2])
    a - 1 = i * tau / (4 * ln[2]) + log_2(3) / 4
    a - 1 + 1 = i * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    a = i * 1 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    Suppose 2^(a - 1) + i * 3^(1 / 4) = 0
    2^(a - 1) + i * 3^(1 / 4) = 0
    2^(a - 1) + i * 3^(1 / 4) - i * 3^(1 / 4) = 0 - i * 3^(1 / 4)
    2^(a - 1) = -i * 3^(1 / 4)
    ln(2^[a - 1]) = ln(-1 * i * 3^[1 / 4])
    ln(2^[a - 1]) = ln(i^2 * i * 3^[1 / 4])
    ln(2^[a - 1]) = ln(i^3 * 3^[1 / 4])
    (a - 1) * ln(2) = ln(i^3 * 3^[1 / 4])
    (a - 1) * ln(2) / ln(2) = ln(i^3 * 3^[1 / 4]) / ln(2)
    (a - 1) * log_2(2) = ln(i^3) / ln(2) + ln(3^[1 / 4]) / ln(2)
    (a - 1) * 1 = ln(e^[i * 3 * tau / 4]) / ln(2) + (1 / 4) * ln(3) / ln(2)
    a - 1 = (i * 3 * tau / 4) * ln(e) / ln(2) + ln(3) / (4 * ln[2])
    a - 1 = (i * 3 * tau / 4) * 1 / ln(2) + ln(3) / (4 * ln[2])
    a - 1 = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4
    a - 1 + 1 = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    a = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    a1 = i * 0 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    a2 = i * 1 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    a3 = i * 2 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
    a4 = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 + 1