Germany | Can you solve this? | Math Olympiad

[@stpat7614]

Including the complex root: 9^x + 9^x = 36 2 * 9^x = 2 * 18 (1 / 2) * 2 * 9^x = (1 / 2) * 2 * 18 9^x = 18 9^x = 2 * 9 9^x / 9 = 2 * 9 / 9 9^x * 9^(-1) = 2 * 9^1 * 9^(-1) 9^(x - 1) = 2 * 9^(1 - 1) 9^(x - 1) = 2 * 9^0 9^(x - 1) = 2 * 1 (3^2)^(x - 1) = 2 3^(2 * [x - 1]) = 2 3^([x - 1] * 2) = 2 (3^[x - 1])^2 = (2^[1 / 2])^2 sqrt([3^(x - 1)]^2) = +/- sqrt([2^(1 / 2)]^2) 3^(x - 1) = +/- 2^(1 / 2) 3^(x - 1) = 2^(1 / 2), or 3^(x - 1) = -2^(1 / 2) Suppose 3^(x - 1) = 2^(1 / 2) 3^(x - 1) = 2^(1 / 2) log(3^[x - 1]) = log(2^[1 / 2]) (x - 1) * log(3) = (1 / 2) * log(2) (x - 1) * log(3) / log(3) = log(2) / (2 * log[3]) (x - 1) * log_3(3) = log_3(2) / 2 (x - 1) * 1 = log_3(2) / 2 x - 1 + 1 = log_3(2) / 2 + 1 x = log_3(2) / 2 + 1 Suppose 3^(x - 1) = -2^(1 / 2) 3^(x - 1) = -2^(1 / 2) 3^(x - 1) = -1 * 2^(1 / 2) 3^(x - 1) = i^2 * 2^(1 / 2) ln(3^[x - 1]) = ln(i^2 * 2^[1 / 2]) (x - 1) * ln(3) = ln(i^2 * 2^[1 / 2]) (x - 1) * ln(3) / ln(3) = ln(i^2 * 2^[1 / 2]) / ln(3) (x - 1) * log_3(3) = ln(i^2) / ln(3) + ln(2^[1 / 2]) / ln(3) (x - 1) * 1 = ln(e^[i * tau / 2]) / ln(3) + ln(2^[1 / 2]) / ln(3) x - 1 = (i * tau / 2) * ln(e) / ln(3) + (1 / 2) * ln(2) / ln(3) x - 1 = (i * tau / 2) * 1 / ln(3) + (1 / 2) * log_3(2) x - 1 = (i * tau) / (2 * ln[3]) + 1 * log_3(2) / 2 x - 1 + 1 = (i * tau) / (2 * ln[3]) + log_3(2) / 2 + 1 x = (i * tau) / (2 * ln[3]) + log_3(2) / 2 + 1 x1 = log_3(2) / 2 + 1 x2 = (i * tau) / (2 * ln[3]) + log_3(2) / 2 + 1

[@AbbasGuclu-jn3po]

9^x=18 x log9=log18 x=log(9×2)/log9 x=(log9+log2)/log9 x=1+log2/log3^2 x=1+1/2 log3(2)

[@adgf1x]

x=log18 base9=1+log2b3/2=(2+log2b3)/2 ans

[@violetadeliu734]

Shkurt: 9 fuqi x=18 . Perdorim log.

[@mehdizangiabadi-iw6tn]

x=.815