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Logarifmik tenglamalar

3 daqiqa o'qish · 10 mashq

Ushbu darsda logarifm belgisi ostida noma’lum qatnashgan tenglamalar — logarifmik tenglamalar haqida so‘z yuritamiz. Logarifm — darajaga ko‘tarishga teskari amal: logab\log_a b degani «aa ni qaysi darajaga ko‘tarsak bb chiqadi?» degan savolga javobdir.

§1. Logarifm va uning xossalari

Ta’rif. a>0, a1, b>0a>0,\ a\ne 1,\ b>0 uchun logab=c    ac=b\log_a b = c \iff a^c = b. Ya’ni logarifm — darajaning ko‘rsatkichi.

Qoida. Asosiy xossalar (x,y>0x,y>0):
loga(xy)=logax+logay\log_a(xy) = \log_a x + \log_a y; logaxy=logaxlogay\quad \log_a \tfrac{x}{y} = \log_a x - \log_a y;
logaxk=klogax\log_a x^k = k\log_a x; alogax=x\quad a^{\log_a x} = x.

Eslatma. Logarifm faqat musbat sondan olinadi. Shuning uchun har bir yechimni aniqlanish sohasiga (ODZ) — logarifm ostidagi ifodalar musbatligiga qarab tekshirish shart.

§2. Eng sodda tenglamalar

Qoida. logaf(x)=c    f(x)=ac\log_a f(x) = c \iff f(x) = a^c (bunda f(x)>0f(x)>0).
logaf(x)=logag(x)    f(x)=g(x)\log_a f(x) = \log_a g(x) \iff f(x) = g(x), hamda f(x)>0f(x)>0.

Misol. log2(x1)=3\log_2 (x-1) = 3 tenglamani yeching.
Yechish. Ta’rif bo‘yicha:

x1=23=8x=9x - 1 = 2^3 = 8 \Rightarrow x = 9

ODZ: x1>0x-1>0, ya’ni x>1x>1 — bajariladi.

Javob. x=9x = 9.

Misol. log5(2x+3)=log5(x+8)\log_5 (2x+3) = \log_5 (x+8) tenglamani yeching.
Yechish. Asoslar teng, logarifm ostini tenglaymiz:

2x+3=x+8x=52x + 3 = x + 8 \Rightarrow x = 5

ODZ: 25+3=13>02\cdot5+3 = 13>0 va 5+8=13>05+8=13>0 — to‘g‘ri.

Javob. x=5x = 5.

§3. Xossalardan foydalanish

Misol. log3x+log3(x2)=1\log_3 x + \log_3 (x-2) = 1 tenglamani yeching.
Yechish. Yig‘indi xossasi: log3[x(x2)]=1\log_3 [x(x-2)] = 1, demak x(x2)=3x(x-2)=3:

x22x3=0(x3)(x+1)=0x=3 yoki x=1x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0 \Rightarrow x=3 \ \text{yoki}\ x=-1

ODZ: x>0x>0 va x2>0x-2>0, ya’ni x>2x>2. x=1x=-1 mos emas.

Javob. x=3x = 3.

Misol. log2(x+2)+log2(x2)=5\log_2 (x+2) + \log_2 (x-2) = 5 tenglamani yeching.
Yechish. log2[(x+2)(x2)]=5x24=25=32\log_2 [(x+2)(x-2)] = 5 \Rightarrow x^2 - 4 = 2^5 = 32:

x2=36x=±6x^2 = 36 \Rightarrow x = \pm 6

ODZ: x>2x>2. Demak x=6x=-6 begona.

Javob. x=6x = 6.

§4. Yangi o‘zgaruvchi kiritish

Misol. log22x3log2x+2=0\log_2^2 x - 3\log_2 x + 2 = 0 tenglamani yeching.
Yechish. t=log2xt = \log_2 x deb belgilaymiz:

t23t+2=0(t1)(t2)=0t=1 yoki t=2t^2 - 3t + 2 = 0 \Rightarrow (t-1)(t-2)=0 \Rightarrow t=1 \ \text{yoki}\ t=2

log2x=1x=2\log_2 x = 1 \Rightarrow x=2;  log2x=2x=4\ \log_2 x = 2 \Rightarrow x=4. Ikkalasi ham x>0x>0.

Javob. x=2; x=4x = 2;\ x = 4.

Savol. Nima uchun logarifmik tenglamada tekshirish ayniqsa muhim? Chunki xossalarni qo‘llaganda (masalan yig‘indini ko‘paytmaga aylantirganda) aniqlanish sohasi kengayishi va begona ildizlar kirib qolishi mumkin.

Eslatma. Doimo boshida ODZ ni yozib qo‘ying: logarifm ostidagi barcha ifodalar musbat, asos musbat va 11 ga teng emas. Bu xatolardan asraydi.

§ Lug‘at

logarifmik tenglama — logarithmic equation
logarifm — logarithm
asos — base
aniqlanish sohasi (ODZ) — domain of definition
logarifm xossalari — properties of logarithms
begona ildiz — extraneous root

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