Ko'rsatkichli tenglama va tengsizliklar

Bu nima va nega kerak

Ko'rsatkichli funksiya butun sonlar o'qida qat'iy monoton (faqat o'suvchi yoki faqat kamayuvchi) bo'lgani sababli, har bir qiymatga argumentning yagona qiymati to'g'ri keladi.

Shu sababli, agar ikkita bir xil asosli daraja teng bo'lsa ($a^{f(x)} = a^{g(x)}$), ularning ko'rsatkichlari ham teng bo'lishi shart: $f(x) = g(x)$.

Ushbu darsda ko'rsatkichli tenglamalarni yechishning 4 ta asosiy usulini va ko'rsatkichli tengsizliklarda asosga qarab ishoraning o'zgarish qoidalarini o'rganamiz.

Asosiy g'oya

1. Ko'rsatkichli tenglamalarni yechish usullari

1. Bir xil asosga keltirish: $a^{f(x)} = a^{g(x)} \iff f(x) = g(x) \quad (a > 0, a \neq 1).$

2. Umumiy ko'paytuvchini qavsdan chiqarish: Darajadagi kichik ko'rsatkich qavsdan tashqariga chiqarilib, qavs ichi songa aylantiriladi: $a^{x+k} + a^x = a^x(a^k + 1).$

3. Yangi o'zgaruvchi kiritish: $A \cdot a^{2x} + B \cdot a^x + C = 0$ tenglamada $t = a^x$ almashtirish olinadi.

   Muhim: Har doim $t > 0$ bo'lishi shart! Manfiy yoki nolga teng $t$ chet ildizdir.

4. Bir jinsli tenglamalar: $A \cdot a^{2x} + B \cdot (ab)^x + C \cdot b^{2x} = 0$ tenglamaning har ikkala tomoni $b^{2x} > 0$ ga bo'linib, $\left(\dfrac{a}{b}\right)^x$ ga nisbatan kvadrat tenglamaga keltiriladi.

2. Ko'rsatkichli tengsizliklar

• $a > 1$ bo'lsa: funksiya o'suvchi, tengsizlik ishorasi saqlanadi: $a^{f(x)} > a^{g(x)} \iff f(x) > g(x).$

• $0 < a < 1$ bo'lsa: funksiya kamayuvchi, tengsizlik ishorasi teskarisiga o'zgaradi: $a^{f(x)} > a^{g(x)} \iff f(x) < g(x).$

Yechilgan misollar

1-misol. Bir xil asosga keltirish

Yeching: a) $27^{x-1} = 9^{x+2}$ b) $\left(\dfrac{1}{2}\right)^{x^2 - 3x} \ge \left(\dfrac{1}{2}\right)^{x + 5}$

Yechish: a) $3$ asosli darajaga keltiramiz: $(3^3)^{x-1} = (3^2)^{x+2} \implies 3^{3x-3} = 3^{2x+4}.$ Asoslar teng, ko'rsatkichlarni tenglashtiramiz: $3x - 3 = 2x + 4 \implies x = 7.$

b) Asos $a = \dfrac{1}{2} < 1$, shuning uchun ishora $\ge$ dan $\le$ ga almashadi: $x^2 - 3x \le x + 5 \implies x^2 - 4x - 5 \le 0 \implies (x - 5)(x + 1) \le 0.$ Oraliqlar usuli bo'yicha: $x \in [-1; 5]$.

Javob: a) $7$; b) $[-1; 5]$.

2-misol. Umumiy ko'paytuvchini qavsdan chiqarish

Tenglamani yeching: $5^{x+1} - 2 \cdot 5^{x-1} = 115.$

Yechish: $5^{x-1}$ ni qavsdan chiqaramiz: $5^{x-1}(5^2 - 2) = 115 \implies 5^{x-1} \cdot 23 = 115.$ Ikkala tomonni $23$ ga bo'lamiz: $5^{x-1} = 5^1 \implies x - 1 = 1 \implies x = 2.$

Javob: $2$.

3-misol. Yangi o'zgaruvchi kiritish

Tenglamani yeching: $9^x - 4 \cdot 3^{x+1} + 27 = 0.$

Yechish: $9^x = (3^x)^2$ va $3^{x+1} = 3 \cdot 3^x$ ekanidan: $(3^x)^2 - 12 \cdot 3^x + 27 = 0.$ $t = 3^x > 0$ belgilash kiritamiz: $t^2 - 12t + 27 = 0 \implies (t - 3)(t - 9) = 0.$ Ildizlar: $t_1 = 3$, $t_2 = 9$. Har ikkisi musbat:

1. $3^x = 3 \implies x_1 = 1$;
2. $3^x = 9 = 3^2 \implies x_2 = 2$.

Javob: $1; 2$.

4-misol. Bir jinsli ko'rsatkichli tenglama

Tenglamani yeching: $3 \cdot 4^x - 5 \cdot 6^x + 2 \cdot 9^x = 0.$

Yechish: Tenglamaning ikkala qismini $9^x > 0$ ga bo'lamiz: $3 \cdot \left(\frac{4}{9}\right)^x - 5 \cdot \left(\frac{6}{9}\right)^x + 2 = 0 \implies 3 \cdot \left(\frac{2}{3}\right)^{2x} - 5 \cdot \left(\frac{2}{3}\right)^x + 2 = 0.$ $t = \left(\dfrac{2}{3}\right)^x > 0$ deb olsak: $3t^2 - 5t + 2 = 0 \implies t_1 = 1, \quad t_2 = \frac{2}{3}.$

1. $\left(\dfrac{2}{3}\right)^x = 1 \implies x_1 = 0$;
2. $\left(\dfrac{2}{3}\right)^x = \dfrac{2}{3} \implies x_2 = 1$.

Javob: $0; 1$.

Mashq

Mustaqil yechish uchun masalalar:

1. $2^{3x - 1} = 32$ tenglamani yeching.
2. $\left(\dfrac{1}{5}\right)^{x^2 - 6} = 25$ tenglamaning ildizlari ko'paytmasini toping.
3. $7^{x+1} + 7^x = 56$ tenglamani yeching.
4. $4^x - 6 \cdot 2^x + 8 = 0$ tenglama ildizlari yig'indisini toping.
5. $3^{x+1} - 3^{x-1} = 24$ tenglamani yeching.
6. $\left(\dfrac{1}{2}\right)^{2x - 1} \le 8$ tengsizlikning eng kichik butun yechimini toping.
7. $(0{,}2)^{x^2 - 2x} > (0{,}2)^3$ tengsizlikning eng katta butun yechimini toping.
8. $4 \cdot 9^x - 7 \cdot 12^x + 3 \cdot 16^x = 0$ tenglamani yeching.

Yechimlar va javoblar

1. $32 = 2^5 \implies 3x - 1 = 5 \implies 3x = 6 \implies x = 2$. Javob: $2$.

2. $25 = 5^2 = \left(\dfrac{1}{5}\right)^{-2} \implies x^2 - 6 = -2 \implies x^2 = 4 \implies x = \pm 2$. Ko'paytma: $2 \cdot (-2) = -4$. Javob: $-4$.

3. $7^x(7 + 1) = 56 \implies 7^x \cdot 8 = 56 \implies 7^x = 7 \implies x = 1$. Javob: $1$.

4. $t = 2^x > 0 \implies t^2 - 6t + 8 = 0 \implies t_1 = 2, t_2 = 4$. $x_1 = 1, x_2 = 2$. Yig'indi: $1 + 2 = 3$. Javob: $3$.

5. $3^{x-1}(3^2 - 1) = 24 \implies 3^{x-1} \cdot 8 = 24 \implies 3^{x-1} = 3 \implies x = 2$. Javob: $2$.

6. $8 = \left(\dfrac{1}{2}\right)^{-3}$. Asos $\dfrac{1}{2} < 1 \implies 2x - 1 \ge -3 \implies 2x \ge -2 \implies x \ge -1$. Eng kichik butun: $-1$. Javob: $-1$.

7. Asos $0{,}2 < 1 \implies x^2 - 2x < 3 \implies (x - 3)(x + 1) < 0 \implies -1 < x < 3$. Eng katta butun son: $2$. Javob: $2$.

8. $16^x$ ga bo'lamiz: $4 \cdot \left(\dfrac{3}{4}\right)^{2x} - 7 \cdot \left(\dfrac{3}{4}\right)^x + 3 = 0$. $t = \left(\dfrac{3}{4}\right)^x \implies 4t^2 - 7t + 3 = 0 \implies t_1 = 1, t_2 = \dfrac{3}{4}$. $x_1 = 0, x_2 = 1$. Javob: $0; 1$.

Keng tarqalgan xatolar

1. Asos $0 < a < 1$ bo'lganda ishorani almashtirishni unutish. ✗ $(0{,}3)^{x-1} > (0{,}3)^2 \implies x - 1 > 2 \implies x > 3$. ✓ Asos $1$ dan kichik bo'lsa funksiya kamayadi, ishora qarama-qarshisiga o'zgaradi: $x - 1 < 2 \implies x < 3$.

2. $t = a^x$ manfiy chiqqanda uni ham yechim deb olish. ✗ $t^2 - 3t - 4 = 0 \implies t_1 = 4, t_2 = -1 \implies 2^x = -1 \implies x = -0{,}5$. ✓ $a^x > 0$ doimo o'rinli. $2^x = -1$ tenglama ildizga ega emas, faqat $2^x = 4 \implies x = 2$ olinadi.

3. Bir xil darajalarni noto'g'ri qo'shish. ✗ $2^x + 2^x = 4^x$ deb yozish. ✓ $2^x + 2^x = 2 \cdot 2^x = 2^{x+1}$. Asoslar faqat ko'paytirilganda darajalar qo'shiladi.

4. Darajani $0$ ga tenglashda asosga e'tibor bermaslik. ✗ $a^{f(x)} = 1$ tenglikda faqat $f(x) = 0$ deb xulosa qilish. ✓ Agar asos ham o'zgaruvchiga bog'liq bo'lsa ($x^{x-2} = 1$), asos $1$ bo'lishi ham mumkin.

DTM uslubidagi test

Ushbu mavzu bo'yicha tayyorlangan 5 ta test savoli platformada interaktiv tarzda ishlaydi: o'z javobingizni tanlang va to'liq yechimni ko'ring.

Keyingi qadam

Ko'rsatkichli tenglamalarda $2^x = 8$ kabi darajalar osongina yechiladi ($x = 3$). Lekin $2^x = 5$ tenglamaning ildizini qanday yozamiz? Keyingi darsda bu savolga javob beruvchi yangi matematik amal — logarifm va uning xossalari bilan tanishamiz.