logarytmy loga: oblicz:

	3x+3−3
a)		= 2
	3x− 3−3

	1		1
b) log		(x+3)+log		(x−2) < −1
	6		6

Basia: (1) 3^x − 3⁻³ ≠ 0 3^x ≠ 3⁻³ x≠ −3 x∊R\{−3} 3^x + 3⁻³ = 2(3^x−3⁻³) 3^x + 3⁻³ = 2*3^x − 2*3⁻³ 3⁻³+2*3⁻³ = 2*3^x − 3^x 3⁻³(1+2) = 3^x(2−1) 3⁻³*3 = 3^x 3^x = 3⁻² x = −2 (2) x+3>0 i x−2>0 x>2 log_1/6(x+3)(x−2) < log_1/66 (x+3)(x−2) > 6 dokończ