Logarytmy naturalne Patryk:

	1		1
ln 3√e, ln		, ln e3, ln1. ln
	e		√e

4 log₂x= log₂81 log_1/2(x−3)+ log_1/2x= −2 log₂(x²−6)=3+log₂(x−1) log₅(5−3x)>1 3tg²x−1≤0 ct2x<3 2sin²x≤1 9^x+3^x−¹⁼4 (√5)^x−³√25=0 (¹₂)^2x−3⁼8

Janek191: ln e = log_e e = 1 więc

	1		1		1
ln 3√e = ln e13 =		ln e =		*1 =
	3		3		3

	1
ln		= ln e−1 = − 1 ln e = − 1*1 = − 1
	e

ln 1 = 0 bo e⁰ = 1 ln e³ = 3 ln e = 3*1 = 3

	1
ln		= ln (√e)−1 = − 1* ln e0,5 = − 1* 0,5 ln e = − 0,5*1 = − 0,5
	√e

Janek191: Np. 4 log₂ x = log₂ 81 ; x > 0 log₂ x⁴ = log₂ 3⁴ x⁴ = 3⁴ x = 3 ==== log_¹2 ( x − 3) + log_¹2 x = − 2 log_0,5 [ ( x − 3)*x] = − 2 log_0,5 0,5

	1
log0,5 ( x2 − 3 x) = log0,5 (		)−2
	2

x² − 3 x = 2² = 4 x² − 3 x − 4 = 0 ( x + 1)*( x − 4) = 0 x = 4 ====