7. \( 2^{x^{2}-3} \cdot 5^{x^{2}-3}=0,01\left(10^{x-1}\right)^{3} \)

Berilgan tenglama: \[ 2^{x^{2}-3} \cdot 5^{x^{2}-3} = 0,01 \left(10^{x-1}\right)^{3} \] **Qadam 1: Tenglamaning ikki tomonini soddalashtirish** Chap tomonni soddalashtiramiz: \[ 2^{x^{2}-3} \cdot 5^{x^{2}-3} = (2 \cdot 5)^{x^{2}-3} = 10^{x^{2}-3} \] O'ng tomonni soddalashtiramiz: \[ 0,01 \left(10^{x-1}\right)^{3} = 0,01 \cdot 10^{3(x-1)} = 0,01 \cdot 10^{3x -3} \] \[ 0,01 = 10^{-2} \], shuning uchun: \[ 10^{-2} \cdot 10^{3x -3} = 10^{3x -5} \] Endi tenglama quyidagicha ko'rinadi: \[ 10^{x^{2}-3} = 10^{3x -5} \] **Qadam 2: Eksponent darajalar tengligini yechish** Agar \(10^a = 10^b\) bo'lsa, unda \(a = b\). Shunday ekan: \[ x^{2} - 3 = 3x -5 \] **Qadam 3: Kvadrat tenglamani yechish** Tenglamani tartibga solamiz: \[ x^{2} - 3x + 2 = 0 \] Bu kvadrat tenglama yechimlarini topamiz: \[ x = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2} \] Demak, yechimlar: \[ x = \frac{3 + 1}{2} = 2 \] \[ x = \frac{3 - 1}{2} = 1 \] **Natija:** \[ x = 1 \quad \text{yoki} \quad x = 2 \]