C+left(1-Pright)^N=1 を解きます| Microsoft 数学ソルバー

$$C+{\left(1-P\right)}^{N}=1$$

C を解く (複素数の解)

$$C=-\left(1-P\right)^{N}+1$$

N を解く (複素数の解)

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

C を解く

$$C=-\left(1-P\right)^{N}+1$$

$$\left(P>1\text{ and }Denominator(N)\text{bmod}2=1\right)\text{ or }\left(P=1\text{ and }N>0\right)\text{ or }P<1$$

N を解く

$$\left\{\begin{matrix}N=\log_{1-P}\left(1-C\right)\text{, }&C<1\text{ and }P\neq 0\text{ and }P<1\\N\in \mathrm{R}\text{, }&\left(P=0\text{ and }C=0\right)\text{ or }\left(P=2\text{ and }C=2\text{ and }Denominator(N)\text{bmod}2=1\text{ and }Numerator(N)\text{bmod}2=1\right)\\N>0\text{, }&P=1\text{ and }C=1\end{matrix}\right.$$