oblicz: 22tysia: a) (¹₈*4^x)^x=2^3x−4 b) 16^x_x−3=4* (^2x₈) ¹_2x−5

dero2005: (¹₈*4^x)^x = 2^3x−4 (2⁻³*2²x) = 2^3x−4 (2^−3+x)^x = 2^3x−4 2^−3x+x2 = 2^3x−4 x²−3x = 3x−4 x² −3x−3x+4 = 0 x²−6x+4 = 0 Δ = 36−16 = 20 √Δ = √20 = 2√5

	6−2√5
x1 =		= 3−√5
	2

	6+2√5
x2 =		= 3+√5
	2

dero2005: 16^x_x−3 = 4*(^2x₈)^1_2x−5 (2⁴)^x_x−3 = 2²*(2^x*2⁻³)^1_2x−5 2^4x_x−3 = 2²*(2^x−3)^1_2x−5 2^4x_x−3 = 2²*2^x−3_2x−5 2^4x_x−3 = 2^{x−3_2x−5+2}

4x		x−3
	=		+ 2
x−3		2x−5

x − 3 ≠ 0 x ≠ 3 2x − 5 ≠ 0 2x ≠ 5 x ≠ ⁵₂ D = R\{⁵₂ , 3} dalej rozwiąż równanie i oblicz x