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]