Równanie Mateusz: 4^x−1+0.5^1−x=0.25^−x−4

Janek191:

1		1
	*4x +		*2x = 4x − 4 / * 4
4		2

(2^x)² + 2*2^x = 4*(2^x)² − 16 3*(2^x)² −2*2^x − 16 = 0 t = 2^x > 0 3 t² −2 t − 16 = 0 Δ = 4 − 4*3*(−16) = 4 + 192 = 196 √Δ = 14

	2 + 14		16		8
t =		=		=
	6		6		3

więc

	8
2x =
	3

	8
log2 2x = log2
	3

x = log₂ 8 − log₂ 3 = 3 − log₂ 3 =========================