Datrys y=1(1+P/100)^x

$$y = 1 ( 1 + \frac { P } { 100 } ) ^ { x }$$

Datrys ar gyfer P (complex solution)

$$P=e^{\frac{Im(x)arg(y)+iRe(x)arg(y)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}-\frac{2\pi n_{1}iRe(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}-\frac{2\pi n_{1}Im(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}}\left(|y|\right)^{\frac{Re(x)-iIm(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}}}$$

$$n_{1}\in \mathrm{Z}$$

Datrys ar gyfer x (complex solution)

$$\left\{\begin{matrix}x=\frac{2\pi n_{1}i}{\ln(\frac{P+100}{100})}+\log_{\frac{P+100}{100}}\left(y\right)\text{, }n_{1}\in \mathrm{Z}\text{, }&y\neq 0\text{ and }P\neq 0\text{ and }P\neq -100\\x\in \mathrm{C}\text{, }&\left(P=-100\text{ and }y=0\right)\text{ or }\left(P=0\text{ and }y=1\right)\end{matrix}\right.$$

Datrys ar gyfer P

$$\left\{\begin{matrix}P=100y^{\frac{1}{x}}-100\text{, }&\left(Numerator(x)\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }y<0\text{ and }y^{\frac{1}{x}}\neq 0\right)\text{ or }\left(y=0\text{ and }x>0\right)\text{ or }\left(y>0\text{ and }x\neq 0\right)\\P=-100y^{\frac{1}{x}}-100\text{, }&\left(y<0\text{ and }Numerator(x)\text{bmod}2=1\text{ and }Numerator(x)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }y^{\frac{1}{x}}\neq 0\right)\text{ or }\left(x\neq 0\text{ and }y>0\text{ and }Numerator(x)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\right)\text{ or }\left(Numerator(x)\text{bmod}2=0\text{ and }y=0\text{ and }x>0\right)\text{ or }\left(y>0\text{ and }x\neq 0\text{ and }y^{\frac{1}{x}}<0\text{ and }Numerator(x)\text{bmod}2=0\right)\\P\neq -100\text{, }&y=1\text{ and }x=0\end{matrix}\right.$$

Datrys ar gyfer x

$$\left\{\begin{matrix}x=\log_{\frac{P+100}{100}}\left(y\right)\text{, }&y>0\text{ and }P\neq 0\text{ and }P>-100\\x\in \mathrm{R}\text{, }&\left(P=0\text{ and }y=1\right)\text{ or }\left(P=-200\text{ and }y=-1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }Numerator(x)\text{bmod}2=1\right)\\x>0\text{, }&P=-100\text{ and }y=0\end{matrix}\right.$$