Ko'rsatkichli va logarifmik masalalar

Bu mavzuda DTM nimani so'raydi

DTM imtihonlarida ko'rsatkichli va logarifmik mavzulardan har yili 2–4 ta savol tushadi:

1. Logarifmlarni hisoblash: asosiy ayniyat $a^{\log_a b} = b$, asosni almashtirish va zanjir qoidasi.
2. Ko'rsatkichli tenglama va tengsizliklar: bir xil asosga keltirish, yangi o'zgaruvchi ($a^x = t > 0$), asos $1$ dan kichik bo'lganda ishoraning o'zgarishi.
3. Logarifmik tenglamalar: aniqlanish sohasi (ODZ) va chet ildizlarni aniqlash.
4. Logarifmik tengsizliklar: asos $0 < a < 1$ bo'lganda ishoraning teskarisiga o'zgarishi va ichki ifodaning musbatligi.

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Asosiy formulalar va usullar

1. Ko'rsatkichli amallar

• $a^{f(x)} \ge a^{g(x)}$ tengsizlikda:
  • agar $a > 1$ bo'lsa: $f(x) \ge g(x)$
  • agar $0 < a < 1$ bo'lsa: $f(x) \le g(x)$ (ishora teskarisiga o'zgaradi!)

2. Logarifm xossalari ($a, b, c > 0, a, c \neq 1$)

• $a^{\log_a b} = b$
• $\log_a(bc) = \log_a b + \log_a c$, $\log_a\left(\dfrac{b}{c}\right) = \log_a b - \log_a c$
• $\log_{a^k}(b^m) = \dfrac{m}{k}\log_a b$
• Asosni almashtirish: $\log_a b = \dfrac{\log_c b}{\log_c a}$, $\log_a b \cdot \log_b c = \log_a c$

3. Logarifmik tengsizliklar

$\log_a f(x) \ge \log_a g(x)$:

• $a > 1 \implies f(x) \ge g(x) > 0$
• $0 < a < 1 \implies 0 < f(x) \le g(x)$

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Yechilgan misollar

1-misol

$4^x - 3 \cdot 2^{x+1} - 16 = 0$ tenglamaning ildizini toping.

A) $2$
B) $3$
C) $4$
D) $8$

Yechimi:
$(2^x)^2 - 6 \cdot 2^x - 16 = 0$.
$t = 2^x > 0$ deb belgilaymiz: $t^2 - 6t - 16 = 0 \implies (t - 8)(t + 2) = 0$

• $t = 8 \implies 2^x = 8 \implies x = 3$.
• $t = -2$ chet ildiz, chunki $2^x > 0$.

To'g'ri javob: B.

Tezroq yo'l: Variantlarni tekshiramiz: $x = 3$ da $4^3 - 3 \cdot 2^4 - 16 = 64 - 48 - 16 = 0$.

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2-misol

Hisoblang: $5^{\log_5 9} + \log_2 3 \cdot \log_3 4 \cdot \log_4 8$.

A) $11$
B) $12$
C) $15$
D) $17$

Yechimi:

1. $5^{\log_5 9} = 9$.
2. Zanjir qoidasidan: $\log_2 3 \cdot \log_3 4 \cdot \log_4 8 = \log_2 8 = 3$.
3. Yig'indi: $9 + 3 = 12$.

To'g'ri javob: B.

Tezroq yo'l: Zanjir qoidasiga ko'ra oraliq asoslar qisqaradi: $\log_2 8 = 3$. Darhol $9 + 3 = 12$.

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3-misol

$\log_2(x^2 - 3x) = 2$ tenglamaning ildizlari ko'paytmasini toping.

A) $4$
B) $3$
C) $-4$
D) $-3$

Yechimi:
$x^2 - 3x = 2^2 = 4 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0$.
Ildizlar: $x_1 = 4, x_2 = -1$.
Ikkala qiymatda ham $x^2 - 3x = 4 > 0$ bajariladi (chet ildiz yo'q).
Ko'paytma: $4 \cdot (-1) = -4$.

To'g'ri javob: C.

Tezroq yo'l: $x^2 - 3x = 4 > 0$ bo'lgani uchun chet ildiz chiqmaydi. Viyet teoremasidan ko'paytma $c = -4$.

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4-misol

$\left(\dfrac{1}{3}\right)^{x^2 - 2x} \ge \dfrac{1}{27}$ tengsizlikning butun yechimlari sonini toping.

A) $3$
B) $4$
C) $5$
D) $6$

Yechimi:
$\left(\dfrac{1}{3}\right)^{x^2 - 2x} \ge \left(\dfrac{1}{3}\right)^3$.
Asos $\dfrac{1}{3} < 1$ bo'lgani uchun ishora teskarisiga o'zgaradi: $x^2 - 2x \le 3 \implies x^2 - 2x - 3 \le 0 \implies (x - 3)(x + 1) \le 0$ Yechim: $x \in [-1; 3]$.
Butun yechimlar: $-1, 0, 1, 2, 3$ (jami $5$ ta).

To'g'ri javob: C.

Tezroq yo'l: Asos $1$ dan kichik bo'lgani sababli $x^2 - 2x \le 3$. Kesmadagi butun sonlar soni: $3 - (-1) + 1 = 5$.

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5-misol

$\log_{0{,}5}(2x - 4) \ge -2$ tengsizlikni yeching.

A) $(-\infty; 4]$
B) $(2; 4]$
C) $[2; 4]$
D) $(2; +\infty)$

Yechimi:

1. Aniqlanish sohasi: $2x - 4 > 0 \implies x > 2$.
2. Asos $0{,}5 < 1$ bo'lgani uchun: $2x - 4 \le (0{,}5)^{-2} = 4 \implies 2x \le 8 \implies x \le 4$
3. Umumiy yechim: $x \in (2; 4]$.

To'g'ri javob: B.

Tezroq yo'l: $x > 2$ bo'lgani uchun $2$ kirmaydi (A va C chiqib ketadi). $x = 4$ yechimga kiradi, demak javob $(2; 4]$.

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Mashq

1. Tenglamani yeching: $3^{2x+1} - 10 \cdot 3^x + 3 = 0$.

Yechimi va javobi

$3t^2 - 10t + 3 = 0 \implies (3t - 1)(t - 3) = 0$.
$3^x = 3 \implies x = 1$; $3^x = \dfrac{1}{3} \implies x = -1$.

Javob: $\{-1; 1\}$

2. Hisoblang: $\log_3 2 \cdot \log_4 9$.

Yechimi va javobi

$\log_4 9 = \log_{2^2} 3^2 = \log_2 3$. Shuning uchun: $\log_3 2 \cdot \log_2 3 = 1$.

Javob: $1$

3. Tenglamani yeching: $\log_3(x+1) + \log_3(x+3) = 1$.

Yechimi va javobi

Aniqlanish sohasi: $x > -1$.
$(x+1)(x+3) = 3 \implies x^2 + 4x = 0 \implies x = 0$ yoki $x = -4$.
$x = -4$ chet ildiz. Faqat $x = 0$ qoladi.

Javob: $0$

4. Tengsizlikni yeching: $2^{x-1} \cdot 3^{x-1} < 36$.

Yechimi va javobi

$6^{x-1} < 6^2 \implies x - 1 < 2 \implies x < 3$.

Javob: $(-\infty; 3)$

5. Hisoblang: $\log_{\sqrt{5}} 25 - \log_2(\log_3 81)$.

Yechimi va javobi

$\log_{5^{1/2}} 5^2 = 4$. $\log_3 81 = 4 \implies \log_2 4 = 2$. Ayirma: $4 - 2 = 2$.

Javob: $2$

6. Tengsizlikning eng kichik butun yechimini toping: $\log_2(x - 5) \ge 3$.

Yechimi va javobi

Aniqlanish sohasi: $x > 5$. $x - 5 \ge 2^3 = 8 \implies x \ge 13$. Eng kichik butun yechim: $13$.

Javob: $13$

7. Tenglamaning ildizlari yig'indisini toping: $x^{\log_3 x} = 9x$.

Yechimi va javobi

$3$ asos bo'yicha logarifmlaymiz: $\log_3^2 x = \log_3 9 + \log_3 x \implies t^2 - t - 2 = 0$.
$t = 2 \implies x = 9$; $t = -1 \implies x = \dfrac{1}{3}$.
Yig'indi: $9 + \dfrac{1}{3} = \dfrac{28}{3}$.

Javob: $\dfrac{28}{3}$

8. Agar $\log_2 3 = a$ bo'lsa, $\log_{12} 18$ ni $a$ orqali ifodalang.

Yechimi va javobi

$\log_{12} 18 = \frac{\log_2 18}{\log_2 12} = \frac{\log_2(2 \cdot 9)}{\log_2(4 \cdot 3)} = \frac{1 + 2\log_2 3}{2 + \log_2 3} = \frac{2a+1}{a+2}$

Javob: $\dfrac{2a+1}{a+2}$

9. Tengsizlikni qanoatlantiruvchi natural sonlar nechta: $0{,}2^{x-3} \ge 25$?

Yechimi va javobi

$5^{3-x} \ge 5^2 \implies 3 - x \ge 2 \implies x \le 1$.
Natural yechim faqat $x = 1$ (jami $1$ ta).

Javob: $1$ ta

10. Tenglamaning ildizlari ko'paytmasini toping: $\lg^2 x - 3\lg x + 2 = 0$.

Yechimi va javobi

$t = \lg x$ desak, $t_1 + t_2 = 3 \implies \lg(x_1 x_2) = 3 \implies x_1 x_2 = 10^3 = 1000$.

Javob: $1000$

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Keng tarqalgan tuzoqlar

1. Ko'rsatkichli tenglamada manfiy qiymatni ildiz deb olish:
   $a^x = t$ belgilash kiritilganda albatta $t > 0$ bo'lishi shart. Agar kvadrat tenglamada $t_2 = -3$ chiqsa, $a^x = -3$ haqiqiy ildizga ega emas.
2. Asos $1$ dan kichik bo'lganda ishorani o'zgartirishni unutish:
   $(0{,}3)^x > (0{,}3)^2$ tengsizlikda $x < 2$ bo'ladi. Asos $0 < a < 1$ bo'lganda ishora qarama-qarshisiga o'zgaradi.
3. Logarifm aniqlanish sohasini (ODZ) tekshirmaslik:
   $\log_a f(x) + \log_a g(x) = \log_a(f(x) \cdot g(x))$ almashtirishida albatta $f(x) > 0$ va $g(x) > 0$ bo'lishi shart, aks holda chet ildizlar kirib qoladi.
4. Logarifmik tengsizlikda ichki ifodaning musbatligini unutish:
   $\log_{0{,}5} f(x) \ge 1 \implies 0 < f(x) \le 0{,}5$. Noldan qat'iy kattalik shartini tashlab ketish xato yechimga olib keladi.