Logarifm va uning xossalari

Bu nima va nega kerak

$2^x = 8$ tenglamani ko'rganda darhol $x = 3$ ekanini aytamiz, chunki $2^3 = 8$. Xuddi shunday, $3^x = 81$ bo'lsa, $x = 4$.

Ammo $2^x = 5$ bo'lsa-chi? $2^2 = 4$ va $2^3 = 8$ bo'lgani uchun $x$ soni $2$ va $3$ oralig'idagi qandaydir irratsional son. Lekin uni qanday aniq yozish mumkin?

Xuddi ko'paytirishga teskari amal bo'lish (bo'linma) va darajaga ko'tarishga teskari amal ildiz chiqarish bo'lganidek, daraja ko'rsatkichini topish amali — logarifm deb ataladi.

Logarifmlar tovush balandligini o'lchash (detsibel), zilzila kuchi (Rixter shkalasi), kimyoviy eritmalarning kislotaliligi (pH) va axborot miqdorini (bit) hisoblashda asosiy apparat hisoblanadi.

Asosiy g'oya

1. Logarifm ta'rifi

Ta'rif: $b$ sonining $a$ asosga ko'ra logarifmi deb, $b$ sonini hosil qilish uchun $a$ asosni ko'tarish kerak bo'lgan daraja ko'rsatkichiga aytiladi.

$\log_a b = c \iff a^c = b \quad (a > 0, \ a \neq 1, \ b > 0)$

Belgilashlar:

• $a$ — logarifm asosi ($a > 0, a \neq 1$);
• $b$ — logarifm ostidagi ifoda ($b > 0$);
• $\log_a b$ — $b$ sonining $a$ asosga ko'ra logarifmi.

Asosiy logarifmik ayniyat: Ta'rifdan bevosita kelib chiqadi: $a^{\log_a b} = b \quad (a > 0, \ a \neq 1, \ b > 0).$

2. Maxsus logarifmlar

Amaliyotda ikkita asos juda keng qo'llaniladi:

1. O'nli logarifm ($a = 10$): $\log_{10} b$ o'rniga $\lg b$ yoziladi. Masalan, $\lg 100 = 2$, $\lg 1000 = 3$.
2. Natural logarifm ($a = e \approx 2{,}718$): $\log_e b$ o'rniga $\ln b$ yoziladi. Masalan, $\ln e = 1$, $\ln e^2 = 2$.

3. Logarifmning asosiy xossalari

Barcha formulalarda $a > 0, a \neq 1$ va argumentlar musbat deb qaraladi:

Xossa nomi	Formula
Bir va asos logarifmi	$\log_a 1 = 0, \quad \log_a a = 1$
Ko'paytmaning logarifmi	$\log_a(x \cdot y) = \log_a x + \log_a y$
Bo'linmaning logarifmi	$\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$
Darajaning logarifmi	$\log_a(x^p) = p \cdot \log_a x$
Asos darajasining logarifmi	$\log_{a^q} x = \dfrac{1}{q} \cdot \log_a x$
Umumiy daraja formulasi	$\log_{a^q}(x^p) = \dfrac{p}{q} \cdot \log_a x$
Yangi asosga o'tish	$\log_a b = \dfrac{\log_c b}{\log_c a} \quad (c > 0, c \neq 1)$
Asos va argument o'rnini almashtirish	$\log_a b = \dfrac{1}{\log_b a}$
Daraja asosi bilan almashtirish	$a^{\log_c b} = b^{\log_c a}$

Yechilgan misollar

1-misol. Ta'rif va asosiy logarifmik ayniyat

Hisoblang: a) $\log_3 \dfrac{1}{27}$ b) $\log_{\sqrt{5}} 25$ c) $5^{2 + \log_5 3}$

Yechish: a) $3$ ning nechanchi darajasi $\dfrac{1}{27}$ ga teng? $\dfrac{1}{27} = 3^{-3}$. Demak, $\log_3 3^{-3} = -3$. b) $\sqrt{5} = 5^{\frac{1}{2}}$ va $25 = 5^2$. $\log_{5^{\frac{1}{2}}} 5^2 = \frac{2}{1/2} \log_5 5 = 4 \cdot 1 = 4.$ c) Daraja xossasi bo'yicha ajratamiz: $5^{2 + \log_5 3} = 5^2 \cdot 5^{\log_5 3} = 25 \cdot 3 = 75.$

Javob: a) $-3$; b) $4$; c) $75$.

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2-misol. Logarifm xossalarini qo'llash

Ifodalarning qiymatini hisoblang: a) $\log_6 12 + \log_6 3$ b) $\log_2 80 - \log_2 5$ c) $\dfrac{\log_3 128}{\log_3 4}$

Yechish: a) Ko'paytma xossasi: $\log_6 12 + \log_6 3 = \log_6(12 \cdot 3) = \log_6 36 = 2$. b) Bo'linma xossasi: $\log_2 80 - \log_2 5 = \log_2\left(\dfrac{80}{5}\right) = \log_2 16 = 4$. c) Yangi asosga o'tish formulasining teskarisi: $\frac{\log_3 128}{\log_3 4} = \log_4 128 = \log_{2^2} 2^7 = \frac{7}{2} = 3{,}5.$

Javob: a) $2$; b) $4$; c) $3{,}5$.

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3-misol. Yangi asosga o'tish va ifodalash

a) Hisoblang: $\log_9 16 \cdot \log_4 27$. b) Agar $\lg 2 = a$ va $\lg 3 = b$ bo'lsa, $\log_5 12$ ni $a$ va $b$ orqali ifodalang.

Yechish: a) Darajalarni oldinga chiqaramiz: $\log_9 16 = \log_{3^2} 2^4 = \frac{4}{2} \log_3 2 = 2 \log_3 2.$ $\log_4 27 = \log_{2^2} 3^3 = \frac{3}{2} \log_2 3.$ Ko'paytiramiz: $2 \log_3 2 \cdot \frac{3}{2} \log_2 3 = 3 \cdot (\log_3 2 \cdot \log_2 3) = 3 \cdot 1 = 3.$

b) $\log_5 12$ ni o'nli logarifmga keltiramiz: $\log_5 12 = \frac{\lg 12}{\lg 5} = \frac{\lg(2^2 \cdot 3)}{\lg(10 / 2)} = \frac{2\lg 2 + \lg 3}{\lg 10 - \lg 2} = \frac{2a + b}{1 - a}.$

Javob: a) $3$; b) $\dfrac{2a + b}{1 - a}$.

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4-misol. Logarifmik tenglamalarda aniqlanish sohasi

Tenglamani yeching: a) $\log_2(x^2 - 1) = 3$ b) $\log_x(3x + 4) = 2$

Yechish: a) Aniqlanish sohasi: $x^2 - 1 > 0$. Ta'rifga ko'ra: $x^2 - 1 = 2^3 = 8 \implies x^2 = 9 \implies x_1 = 3, \quad x_2 = -3.$ Ikkala qiymat ham $x^2 - 1 > 0$ shartini qanoatlantiradi ($9 - 1 = 8 > 0$).

b) Aniqlanish sohasi: asos $x > 0, x \neq 1$ va argument $3x + 4 > 0$. Ta'rifga ko'ra: $x^2 = 3x + 4 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0.$ Ildizlar: $x = 4$ va $x = -1$. $x = -1$ shartga zid ($x > 0$), demak, chet ildiz. Faqat $x = 4$ qoladi.

Javob: a) $\pm 3$; b) $4$.

Mashq

Mustaqil yechish uchun masalalar:

1. Hisoblang: $\log_2 24 - \log_2 3$.
2. Qiymatini toping: $4^{\log_2 5}$.
3. Hisoblang: $\log_8 125 \cdot \log_5 2$.
4. $\log_3(2x + 1) = 3$ tenglamani yeching.
5. Hisoblang: $\lg 25 + \lg 4$.
6. Agar $\log_7 2 = a$ bo'lsa, $\log_7 28$ ni $a$ orqali ifodalang.
7. Qiymatini toping: $9^{1 - \log_3 2}$.
8. $\log_x 64 = 3$ tenglamani yeching.

Yechimlar va javoblar

1. $\log_2\left(\dfrac{24}{3}\right) = \log_2 8 = 3$. Javob: $3$.

2. $4^{\log_2 5} = (2^2)^{\log_2 5} = (2^{\log_2 5})^2 = 5^2 = 25$. Javob: $25$.

3. $\log_8 125 = \log_{2^3} 5^3 = \dfrac{3}{3}\log_2 5 = \log_2 5$. Natija: $\log_2 5 \cdot \log_5 2 = 1$. Javob: $1$.

4. Aniqlanish sohasi: $2x + 1 > 0 \implies x > -0{,}5$. $2x + 1 = 3^3 = 27 \implies 2x = 26 \implies x = 13$. Javob: $13$.

5. $\lg(25 \cdot 4) = \lg 100 = 2$. Javob: $2$.

6. $\log_7 28 = \log_7(4 \cdot 7) = \log_7 2^2 + \log_7 7 = 2\log_7 2 + 1 = 2a + 1$. Javob: $2a + 1$.

7. $9^{1 - \log_3 2} = \dfrac{9}{9^{\log_3 2}} = \dfrac{9}{(3^{\log_3 2})^2} = \dfrac{9}{2^2} = \dfrac{9}{4} = 2{,}25$. Javob: $2{,}25$ (yoki $\dfrac{9}{4}$).

8. Aniqlanish sohasi: $x > 0, x \neq 1$. $x^3 = 64 \implies x = 4$. Javob: $4$.

Keng tarqalgan xatolar

1. Logarifmlar yig'indisini argumentlar yig'indisi deb hisoblash. ✗ $\log_a(x + y) = \log_a x + \log_a y$ deb xato yozish. ✓ Argumentlar ko'paytirilganda logarifmlar qo'shiladi: $\log_a(x \cdot y) = \log_a x + \log_a y$. Yig'indining logarifmi soddalashtirilmaydi.

2. Asos darajasini chiqarishda maxrajga tushirishni unutish. ✗ $\log_{a^q} b = q \log_a b$ deb yozish. ✓ Argument darajasi to'g'ridan-to'g'ri ko'paytiriladi ($p$), asos darajasi esa maxrajga bo'linadi: $\log_{a^q} b = \dfrac{1}{q} \log_a b$.

3. Manfiy sonning logarifmini mavjud deb o'ylash. ✗ $\log_2(-4) = -2$ deb hisoblash ($2^{-2} = \dfrac{1}{4} \neq -4$). ✓ Logarifm ostidagi ifoda doimo qat'iy musbat bo'lishi shart ($b > 0$). Manfiy sonning logarifmi mavjud emas.

4. Aniqlanish sohasini (chet ildizlarni) tekshirmaslik. ✗ $\log_x 9 = 2 \implies x^2 = 9 \implies x = \pm 3$. ✓ Asos $x > 0$ va $x \neq 1$ bo'lishi kerak. Shuning uchun $x = -3$ chet ildiz bo'lib, faqat $x = 3$ olinadi.

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Keyingi qadam

Logarifm va uning asosiy hisoblash qoidalari bilan qurollandik. Keyingi darsda argumenti logarifm ichida bo'lgan logarifmik funksiya, uning grafigi va xossalari bilan tanishamiz.