Logarytmy :) Koczer: 1) log₁₂18*log₂₄54+5(log₁₂18−log₂₄54)=1 2) 1+log₃7*log₇5*log₅4=log₃12

@Basia: Pomagam

@Basia: ad.2

	log57
log37 =
	log53

	1
log75 =
	log57

	log57		1		log54
L = 1 +		*		*log54 = 1 +
	log53		log57		log53

	log34
log54 =
	log35

	log33		1
log53 =		=
	log35		log35

	log34
L = 1 +		*log35 = 1 + log34 =
	log35

log₃3 + log₃4 = log₃(3*4) = log₃12 = P nad (1) myślę

@Basia: Policzyłam i zaraz napiszę, ale rachunki są dość wredne. Może ktoś znajdzie prostszy sposób.

@Basia: 12 = 3*2² 18 = 2*3² log₁₂18 = a ⇔ 12^a = 18 ⇔ (3*2²)^a = 2*3² ⇔ 3^a*2^2a = 2*3² ⇔

22a		32
	=		⇔ 22a−1 = 32−a ⇔ (po zlogarytmowaniu log2)
2		3a

2a−1 = (2−a)*log₂3 stąd: 2a +a*log₂3 = 2*log₂3+1 a(2+log₂3) = 2log₂3+1

	2log23+1
a =
	2+log23

24 = 3*2³ 54 = 2*3³ log₂₄54 = b ⇔ 24^b = 54 ⇔ (3*2³)^b = 2*3³ ⇔ 3^b*2^3b = 2*3³ ⇔

23b		33
	=		⇔ 23b−1 = 33−b ⇔ (po zlogarytmowaniu log2)
2		3b

3b−1 = (3−b)*log₂3 3b + b*log₂3 = 3*log₂3 + 1 b(3+log₂3) = 3log₂3+1

	3log23+1
b =
	3+log23

podstawiam: t=log₂3 wtedy:

	2t+1
log1218 = a =
	2+t

	3t+1
log2454 = b =
	3+t

L = a*b + 5(a−b) =

2t+1		3t+1		2t+1		3t+1
	*		+ 5(		−		) =
2+t		3+t		2+t		3+t

(2t+1)(3t+1)		(2t+1)(3+t) − (3t+1)(2+t)
	+ 5*		=
(2+t)(3+t)		(2+t)(3+t)

6t2+2t+3t+1+5[ 6t+2t2+3+t − (6t+3t2+2+t)]
	=
6+2t+3t+t2

6t2+5t+1+5[ 2t2+7t+3−3t2−7t−2]
	=
t2+5t+6

6t2+5t+1+5(−t2+1)
	=
t2+5t+6

t2+5t+6
	= 1
t2+5t+6

Koczer: Dziekuje bardzo