odp. 3Silnia&6: 55) log₂ 24 = a, obliczyc log₂ 3 log₂ 24 = a log₂ (8*3) = a log₂ 8 + log₂ 3 =a 3 + log₂ 3 = a ⇔ log₂ 3 = a − 3

3Silnia&6: log₂ 8 = log₂ 2³ = 3

3Silnia&6: log₂ 36 = log₂ (4 * 9 ) = log₂ 4 + log₂ 9 = log₂ 2² + log₂ 3² = 2 + 2log₂ 3 = 2 + 2(a−3) = 2 + 2a − 6 = 2a − 4

3Silnia&6: 1) log₂ 3 = a. oblicz log₂ 6 . 2) log₃ 7 = b. oblicz log₃ 3087

3Silnia&6: log₂ 6 = log₂ (2*3) = log₂ 2+ log₂ 3 = 1 + a a bo log₂ 3 to a i log_x x = 1 log_x xⁿ = n

3Silnia&6: 2) log₂ 3 = a . Oblicz log₃ 6

	logc b
loga b =		( wzor na zamiane podstaw )
	loga b

dla c = b

	logb b		1
loga b =		= loga b =
	logb a		logb a

3Silnia&6: w naszym przypadku ...

	log2 6
log3 6 =
	log2 3

zamieniam na log o podstawie z 2, bo w zadaniu jest PODANY logarytm o PODSTAWIE 2 (log₂ 3 = a)

	log2 6
log3 6 =
	log2 3

log2 6		log2 (2*3)		log2 2 + log2 3
	=		=
log2 3		log2 3		log2 3

podstawiam log₂ 3 jako (a) i za log₂ 2 (jeden)

	log2 2 + log2 3		1+a
log3 6 ==		=
	log2 3		a

3Silnia&6: log₂ 3 = a log₂ 5 = b oblicz: a) log 3 5 b) log₂ 90

3Silnia&6: Znasz wzor na zamiane podstaw z zadania mamy: log₂ 3 = a log₂ 5 = b mamy obliczyc log₃5

	logcb
logb a =
	logca

Wiec

	log2 5		a
log35 =		=
	log2 3		b

3Silnia&6: b/a xD

3Silnia&6: Pamiętasz to ?

3Silnia&6: log_a a^b = b np. log₂ 2⁴ = 4 log₃ 9 = log₃ 3² = 2 log₅ 125 = log₅ 5³ = 3 log₃ 3¹0231 = 10231

	1
log7		= log7 7−1 = −1
	7

	1
log7 √7 = log7 71/2 =
	2

	1
log5		= log5 5−2 = −2
	25

log₂ 2√2 = log₂ (2¹ * 2^1/2 ) = log₂ 2^3/2 = 3/2 log₃ 3√27 = log₃ (3*3*√3) = log₃ (3² * 3¹{2}) = log₃ 3^2+1/2 = 2 + 1/2 = 5/2 log₃ x = 8 wiec 3⁸ = x log₂ x = 4 ⇔ x = 2⁴ , x = 16 log₃ (x+1) = 2 ⇔ x + 1 = 3² ⇔ x + 1 = 9 , x =8 log_x 3 = 12 wiec x¹2 = 3 => x = p12{3} log_x 10 = 2 <=> x² = 10 <=> x = √10 , bo x > 0 i x ≠ 1

3Silnia&6:

	1
log2 x = −1 ⇔ 2−1 = x ⇔		= x
	2

Хпї: 5 * 2 x <= 2 x + 7

Анна Богач: 5 * 2 x <= 2 x + 7