6y=9^x-3^x+2+1 eching

$$6 y = 9 ^ { x } - 3 ^ { x + 2 } + 1$$

y uchun yechish

$$y=\frac{9^{x}-9\times 3^{x}+1}{6}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Shartlarni qayta saralash.

$$6y=-3^{x+2}+9^{x}+1$$

Tenglama standart shaklda.

$$6y=1+9^{x}-3^{x+2}$$

Ikki tarafini 6 ga bo‘ling.

$$\frac{6y}{6}=\frac{9^{x}-9\times 3^{x}+1}{6}$$

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

$$y=\frac{9^{x}-9\times 3^{x}+1}{6}$$

-9\times 3^{x}+9^{x}+1 ni 6 ga bo'lish.

$$y=\frac{9^{x}}{6}-\frac{3\times 3^{x}}{2}+\frac{1}{6}$$

x uchun yechish (complex solution)

$$\left\{\begin{matrix}\\x=\log_{3}\left(\frac{\sqrt{24y+77}+9}{2}\right)+\frac{2\pi n_{2}i}{\ln(3)}\text{, }n_{2}\in \mathrm{Z}\text{, }&\text{unconditionally}\\x=\log_{3}\left(\frac{-\sqrt{24y+77}+9}{2}\right)+\frac{2\pi n_{1}i}{\ln(3)}\text{, }n_{1}\in \mathrm{Z}\text{, }&y\neq \frac{1}{6}\end{matrix}\right,$$

x uchun yechish

$$\left\{\begin{matrix}x=\log_{3}\left(\frac{\sqrt{24y+77}+9}{2}\right)\text{, }&y\geq -\frac{77}{24}\\x=\log_{3}\left(\frac{-\sqrt{24y+77}+9}{2}\right)\text{, }&y\geq -\frac{77}{24}\text{ and }y<\frac{1}{6}\end{matrix}\right,$$