Germany | Can you solve this? | Math Olympiad

[@stpat7614]

Including the complex roots: 5^a * 5^a * 5^a = 60 5^(a + a + a) = (60^[1 / 3])^3 5^(3 * a) = ([2^2 * 3 * 5]^[1 / 3])^3 5^(a * 3) = ([2^2]^[1 / 3] * 3^[1 / 3] * 5^[1 / 3])^3 (5^a)^3 = (2^[2 * (1 / 3)] * 3^[1 / 3] * 5^[1 / 3])^3 (5^a)^3 = (2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3])^3 Let x = 5^a, and y = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3) x^3 = y^3 x^3 - y^3 = y^3 - y^3 x^3 - y^3 = 0 (x - y)(x^2 + xy + y^2) = 0 (x - y)(1 * x^2 + y * x + y^2) = 0 x - y = 0, or 1 * x^2 + y * x + y^2 = 0 Suppose x - y = 0 x - y = 0 x - y + y = 0 + y x = y Remember, x = 5^a, and y = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3) 5^a = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3) log(5^a) = log(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3]) a * log(5) = log(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3]) a * log(5) / log(5) = log(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3]) / log(5) a * log_5(5) = log_5(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3]) a * 1 = log_5(2^[2 / 3]) + log_5(3^[1 / 3]) + log_5(5^[1 / 3]) a = (2 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * log_5(5) a = 2 * log_5(2) / 3 + 1 * log_5(3) / 3 + (1 / 3) * 1 a = 2 * log_5(2) / 3 + log_5(3) / 3 + (1 / 3) Suppose 1 * x^2 + y * x + y^2 = 0 1 * x^2 + y * x + y^2 = 0 x = (-y +/- sqrt[y^2 - 4 * 1 * y^2]) / (2 * 1) x = (-y +/- sqrt[1 * y^2 - 4 * y^2]) / (2) x = (-y +/- sqrt[(1 - 4) * y^2]) / 2 x = (-y +/- sqrt[(-3) * y^2]) / 2 x = (-y +/- sqrt[(-1) * 3 * y^2]) / 2 x = (-y +/- sqrt[-1] * sqrt[3] * sqrt[y^2]) / 2 x = (-y +/- i * 3^[1 / 2] * y) / 2^(3 / 3) x = 2^(-3 / 3) * y * (-1 +/- i * 3^[1 / 2] * 1) Remember, x = 5^a, and y = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3) 5^a = 2^(-3 / 3) * 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 +/- i * 3^[1 / 2]) 5^a = 2^([-3 / 3] + [2 / 3]) * 3^(1 / 3) * 5^(1 / 3) * (-1 +/- i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 +/- i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 + i * 3^[1 / 2]) or 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 - i * 3^[1 / 2]) Suppose 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 + i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 + i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (i^2 + i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * i * (i + 1 * 3^[1 / 2]) ln(5^a) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i * [i + 3^(1 / 2)]) a * ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i * [i + 3^(1 / 2)]) a * ln(5) / ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i * [i + 3^(1 / 2)]) / ln(5) a * log_5(5) = ln(2^[-1 / 3]) / ln(5) + ln(3^[1 / 3]) / ln(5) + ln(5^[1 / 3]) / ln(5) + ln(i) / ln(5) + ln(i + 3^[1 / 2]) / ln(5) a * 1 = log_5(2^[-1 / 3]) + log_5(3^[1 / 3]) + log_5(5^[1 / 3]) + ln(e^[i * tau / 4]) / ln(5) + log_5(i + 3^[1 / 2]) a = (-1 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * log_5(5) + (i * tau / 4) * ln(e) / ln(5) + log_5(i + 3^[1 / 2]) a = (-1 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * 1 + (i * tau / 4) * 1 / ln(5) + log_5(i + 3^[1 / 2]) a = -1 * log_5(2) / 3 + 1 * log_5(3) / 3 + (1 / 3) + i * tau / (4 * ln[5]) + log_5(i + 3^[1 / 2]) a = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (4 * ln[5]) + log_5(i + 3^[1 / 2]) Suppose 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 - i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 - i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1) * (1 + i * 3^[1 / 2]) 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * i^2 * (1 + i * 3^[1 / 2]) ln(5^a) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i^2 * [1 + i * 3^(1 / 2)]) a * ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i^2 * [1 + i * 3^(1 / 2)]) a * ln(5) / ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i^2 * [1 + i * 3^(1 / 2)]) / ln(5) a * log_5(5) = ln(2^[-1 / 3]) / ln(5) + ln(3^[1 / 3]) / ln(5) + ln(5^[1 / 3]) / ln(5) + ln(i^2) / ln(5) + ln(1 + i * 3^[1 / 2]) / ln(5) a * 1 = ln(2^[-1 / 3]) / ln(5) + ln(3^[1 / 3]) / ln(5) + ln(5^[1 / 3]) / ln(5) + ln(e^[i * tau / 2]) / ln(5) + ln(1 + i * 3^[1 / 2]) / ln(5) a = (-1 / 3) * ln(2) / ln(5) + (1 / 3) * ln(3) / ln(5) + (1 / 3) * ln(5) / ln(5) + (i * tau / 2) * ln(e) / ln(5) + ln(1 + i * 3^[1 / 2]) / ln(5) a = (-1 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * log_5(5) + (i * tau / 2) * 1 / ln(5) + log_5(1 + i * 3^[1 / 2]) a = -1 * log_5(2) / 3 + 1 * log_5(3) / 3 + 1 * log_5(5) / 3 + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2]) a = -log_5(2) / 3 + log_5(3) / 3 + log_5(5) / 3 + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2]) a = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2]) a1 = 2 * log_5(2) / 3 + log_5(3) / 3 + (1 / 3) a2 = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (4 * ln[5]) + log_5(i + 3^[1 / 2]) a3 = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2])

[@louisvanrijn3964]

Please comment: 3 . 5 ^a = 60 ; 5 ^a =20 = 4 .5 ; 5log5 ^a = 5log (4 . 5) ; a = 5log4 +1 ??? OK??

[@ЮрийФетисов-с4ъ]

Зачем забивать мозги,не нужными задачами????это нигде и никогда не пригодится

[@AlexanderSemashkevich]

Тебе - не пригодится. Да твои мозги не забить, даже если бы хотелось: у тебя в двух коротких предложениях - несколько грамматических ошибок.

[@AlexanderSemashkevich]

x=log60/3log5=(log5+log4+log3)/3log5=⅓+(2log2+log3)/3log5 x≈⅓+(2×0.301+0.477)/(3×0.699)≈1778/2097≈0.848 Deviation is about 0.0001 5^(3×0.848)≈60.004