y=ab^x eching

$$y = a b ^ { x }$$

a uchun yechish (complex solution)

$$\left\{\begin{matrix}a=\frac{y}{b^{x}}\text{, }&x=0\text{ or }b\neq 0\\a\in \mathrm{C}\text{, }&y=0\text{ and }b=0\text{ and }x\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$ab^{x}=y$$

Tenglama standart shaklda.

$$b^{x}a=y$$

Ikki tarafini b^{x} ga bo‘ling.

$$\frac{b^{x}a}{b^{x}}=\frac{y}{b^{x}}$$

b^{x} ga bo'lish b^{x} ga ko'paytirishni bekor qiladi.

$$a=\frac{y}{b^{x}}$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{y}{b^{x}}\text{, }&b>0\text{ or }\left(Denominator(x)\text{bmod}2=1\text{ and }b<0\right)\\a\in \mathrm{R}\text{, }&y=0\text{ and }b=0\text{ and }x>0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$ab^{x}=y$$

Tenglama standart shaklda.

$$b^{x}a=y$$

Ikki tarafini b^{x} ga bo‘ling.

$$\frac{b^{x}a}{b^{x}}=\frac{y}{b^{x}}$$

b^{x} ga bo'lish b^{x} ga ko'paytirishni bekor qiladi.

$$a=\frac{y}{b^{x}}$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}b=e^{-\frac{2\pi n_{1}iRe(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}-\frac{2\pi n_{1}Im(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}+\frac{arg(\frac{y}{a})Im(x)+iarg(\frac{y}{a})Re(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}}\times \left(\frac{|y|}{|a|}\right)^{\frac{Re(x)-iIm(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}}\text{, }n_{1}\in \mathrm{Z}\text{, }&a\neq 0\\b\in \mathrm{C}\text{, }&y=0\text{ and }a=0\end{matrix}\right,$$