logarytmy rozszerzenie pacman7c3: x^3+2logx=10x^1−logx

Basia: y = logx ⇔ 10^y = x (10^y)^3+2y = 10*(10^y)^1−y 10^3y+2y2 = 10*10^y−y2 10^3y+2y2 = 10^1+y−y2 3y+2y² = 1+y − y² 3y² + 2y − 1 = 0 Δ = 4 − 4*3*(−1) = 16

	−2−4
y1 =		= −1
	6

	−2+4		1
y2 =		=
	6		3

logx = −1 ⇒ x=10⁻¹ = 0,1 logx = ¹₃ ⇒ x=10^1/3 = ³√10

pigor: ... , rozwiązań szukam wśród liczb x∊R₊, więc np. tak : x^3+2logx = 10x^1−logx ⇔ log(x^3+2logx) = log(10x^1−logx} ⇔ ⇔ (3+2logx)logx = log10+logx^1−logx ⇔ 3logx+2log²x = 1+(1−logx)logx ⇔ ⇔ 3logx+2log²x = 1+logx−log²x ⇔ 3log²x+2logx−1= 0 ⇔ ⇔ 3log²x+3logx−logx−1= 0 ⇔ 3logx (logx+1)−1 (logx+1)= 0 ⇔ ⇔ (logx+1) (3logx−1)= 0 ⇔ logx+1=0 lub 3logx−1= 0 ⇔ ⇔ logx=−1 lub logx=¹₃ ⇔ x=10⁻¹ lub x=10 ^1₃ ⇔ x∊{¹₁₀,³√10}

pacman7c3: Bardzo dobre rozwiązania . Brawo i dzięki .