Germany | Can you solve this? | Math Olympiad

[@stpat7614]

Including the complex root: 4^(x + 2) = 20 (2^2)^(x + 2) = (sqrt[20])^2 2^(2 * [x + 2]) = (sqrt[2^2 * 5])^2 2^([x + 2] * 2) = (sqrt[2^2] * sqrt[5])^2 (2^[x + 2])^2 = ([2^2]^[1 / 2] * 5^[1 / 2])^2 (2^[x + 2])^2 = (2^[2 * (1 / 2)] * 5^[1 / 2])^2 (2^[x + 2])^2 = (2^1 * 5^[1 / 2])^2 (2^x + 2^2)^2 = (2 * 5^[1 / 2])^2 Let a = 2^x * 2^2, and b = 2 * 5^[1 / 2] a^2 = b^2 a^2 - b^2 = b^2 - b^2 a^2 - b^2 = 0 (a - b)(a + b) = 0 (2^x * 2^2 - 2 * 5^[1 / 2])(2^x * 2^2 + 2 * 5^[1 / 2]) = 0 2 * (2^x * 2^1 - 2^0 * 5^[1 / 2]) * 2 * (2^x * 2^1 + 2^0 * 5^[1 / 2]) = 0 (2 * 2) * (2^x * 2 - 1 * 5^[1 / 2])(2^x * 2 + 1 * 5^[1 / 2]) = 0 (1 / 2^2) * 2^2 * (2^x * 2 - 5^[1 / 2])(2^x * 2 + 5^[1 / 2]) = (1 / 2^2) * 0 (2^x * 2 - 5^[1 / 2])(2^x * 2 + 5^[1 / 2]) = 0 Suppose 2^x * 2 - 5^(1 / 2) = 0 2^x * 2 - 5^(1 / 2) = 0 2^x * 2 - 5^(1 / 2) + 5^(1 / 2) = 0 + 5^(1 / 2) 2^x * 2 = 5^(1 / 2) 2^x * 2 / 2 = 5^(1 / 2) / 2 2^x * 1 = 5^(1 / 2) / 2 log(2^x) = log(5^[1 / 2] / 2) x * log(2) = log(5^[1 / 2] / 2) x * log(2) / log(2) = log(5^[1 / 2] / 2) / log(2) x * log_2(2) = log_2(5^[1 / 2] / 2) x * 1 = log_2(5^[1 / 2]) - log_2(2) x = (1 / 2) * log_2(5) - 1 x = log_2(5) / 2 - 1 x1 = log_2(5) / 2 - 1 Suppose 2^x * 2 + 5^(1 / 2) = 0 2^x * 2 + 5^(1 / 2) = 0 2^x * 2 + 5^(1 / 2) - 5^(1 / 2) = 0 - 5^(1 / 2) 2^x * 2 = -5^(1 / 2) 2^x * 2 / 2 = -5^(1 / 2) / 2 2^x * 1 = -5^(1 / 2) / 2 ln(2^x) = ln(-5^[1 / 2] / 2) x * ln(2) = ln(-5^[1 / 2] / 2) x * ln(2) / ln(2) = ln(-5^[1 / 2] / 2) / ln(2) x * log_2(2) = ln(-1 * 5^[1 / 2] / 2) / ln(2) x * 1 = ln(-1) / ln(2) + ln(5^[1 / 2] / 2) / ln(2) x = ln(e^[i * tau / 2]) / ln(2) + log_2(5^[1 / 2] / 2) x = (i * tau / 2) * ln(e) / ln(2) + log_2(5^[1 / 2]) - log_2(2) x = (i * tau / 2) * 1 / ln(2) + (1 / 2) * log_2(5) - 1 x = i * tau / (2 * ln[2]) + log_2(5) / 2 - 1 x2 = i * tau / (2 * ln[2]) + log_2(5) / 2 - 1 {x1, x2} = { [log_2(5) / 2 - 1], [i * tau / (2 * ln[2]) + log_2(5) / 2 - 1] }

[@AlexanderSemashkevich]

(x+2)log4=log20 x=(log4+log5)/log4-2=1+log5/log4-2=log5/2log2-1 x≈0.699/(2×0.301)-1≈97/602≈0.161 Deviation is about 0.0002 4^(0.161+2)≈20.001

[@MasterTMathsClass]

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[@dietrichschoen7340]

4^(x)*4^(2)=20 4^(x)=5/4 x=ln(5/4)/ln(4) x=0,161 😊

[@andresrebolledobanquet1924]

4x + 8 = 20 4x = 20 - 8 4x = 12 x = log 12 / log 4 x = 1.79248 4^1.79248 = 12