0 \\ \\ 3t + \dfrac{18}{t} = 29 \\ \\ 3t^2 - 29t + 18 = 0 \\ 3t^2 - 27t - 2t + 18 = 0 \\ 3t(t - 9) - 2(t - 9) = 0 (3t - 2)(t - 9) = 0 \\ t = 1,5 \\ t = 9 " alt=" 3^{x + 1} + \dfrac{18}{3^{x}} =29 \\ \\ 3 \cdot 3^{x} + \dfrac{18}{3^{x}} =29 \\ \\ t = 3^{x}, \ t > 0 \\ \\ 3t + \dfrac{18}{t} = 29 \\ \\ 3t^2 - 29t + 18 = 0 \\ 3t^2 - 27t - 2t + 18 = 0 \\ 3t(t - 9) - 2(t - 9) = 0 (3t - 2)(t - 9) = 0 \\ t = 1,5 \\ t = 9 " align="absmiddle" class="latex-formula">
Обратная замена:
3ˣ = 1,5
x = log₃(1,5)
3ˣ = 9
x = 2
Ответ: x = log₃(1,5); 2.