zadanie 1 a) [latex]16^x - 6*4^x + 8 = 0\ (4^x)^2- 6*4^x + 8 = 0\ t = 4^x > 0\ t^2 - 6t + 8 = 0\ (t - 4)(t - 2) = 0\ 4^x = 4 vee 4^x = 2\ x = 1 vee x = frac{1}{2}[/latex] b) [latex]2^x < frac{1}{2sqrt{2}} = frac{1}{2^{1 + frac{1}{2}}} = 2^{- frac{3}{2}} \ x < - frac{3}{2}[/latex] zadanie 2 a) dziedzina y > 2 [latex]log_2(y + 2) + log_2(y - 2) = 5\ log_2(y + 2)(y - 2) = 5\ 2^5 = (y + 2)(y - 2)\ 32 = y^2 - 4\ y^2 = 36\ y = 6 vee y = - 6\ y > 2 Rightarrow y = 6[/latex] b) dziadzina x > 2,5 [latex]log_{frac{1}{2}}(2x - 5) > - 4\ log_{frac{1}{2}}(2x - 5) > log_{frac{1}{2}}16\ 2x - 5 < 16\ 2,5 < x < 10,5[/latex] jak masz pytania to pisz na pw
z.1 a) 16^x - 6*4^x + 8 = 0 (4²)^x - 6*4^x + 8 = 0 (4^x)² - 6 *4^x + 8 = 0 y = 4^x y² - 6 y + 8 = 0 Δ = 36 - 4*1*8 = 36 - 32 = 4 √Δ = 2 y = [6 - 2]/2 = 2 lub y = [6 +2]/2 = 4 zatem 4^x = 2 lub 4^x = 4 x = 1/2 lub x = 1 ======================= b) 2^x < 1/(2√2) 2^x < 1/[2^(3/2)] 2^x < 2^(- 3/2) x < - 3/2 z.2 a) log₂ (y +2) = log₂ (y -2) = 5 y - 2 > 0 --> y > 2 log₂(y +2)*(y -2) = 5 2⁵ = (y +2)*(y -2) 32 = y² - 4 y² = 32 + 4 = 36 y = - 6 < 2 ( odpada ) y = 6 ============ spr. log₂ (6+2) + log(6 -2) = log₂8*4 = log₂ 32 = 5 b) log½ (2x - 5) > -4 ; 2x - 5 > 0 --> x > 2,5 log½ (2x - 5) > log½ 16 2x - 5 < 16 2x < 16 + 5 = 21 x < 10,5 oraz x > 2,5 czyli x ∈ ( 2,5 ; 10,5) ===========================================
