[latex]2*3^{2x+4} - 245*3^{x}+3 \leq 0[/latex]

Решение 2*3^(2x + 4) - 245*(3^x) + 3 ≤ 0 2*81*(3^2x) - 245*(3^2x) + 3 ≤ 0 162*(3^2x) - 245*(3^x) + 3 ≤ 0 3^x = t 162t² - 245t + 3 ≤ 0 D = 245² - 4*162*3 =  60025 - 1944 = 58081 t₁ = (245 - 241)/324 t₁ = 4/324 t₁ = 1/81 t₂ = (245 + 241)/324 t₂ = 486/324 t₂ = 1,5 3^x = 1/81 3^x = 3⁻⁴ x₁ = - 4 3^x = 3/2 log₃ (3^x) = log₃ (3/2) x log₃ 3 = log₃ 3 - log₃ 2 x₂ = 1 - log₃ 2 x ∈[ - 4; 1 - log₃ 2] или log₃ 2 ≈ 0,63 x₂ ≈ 1 - 0,63 = 0,37 x∈ [- 4 ; 0,37]