Rezolvați 12-2q^N(1/3)=K

$$12 - 2 q ^ { N } ( 1 / 3 ) = K$$

Rezolvați pentru K (complex solution)

$$K=-\frac{2q^{N}}{3}+12$$

Rezolvați pentru N (complex solution)

$$\left\{\begin{matrix}N=\frac{2\pi n_{1}i}{\ln(q)}+\log_{q}\left(-\frac{3K}{2}+18\right)\text{, }n_{1}\in \mathrm{Z}\text{, }&K\neq 12\text{ and }q\neq 1\text{ and }q\neq 0\\N\in \mathrm{C}\text{, }&\left(q=0\text{ and }K=12\right)\text{ or }\left(q=1\text{ and }K=\frac{34}{3}\right)\end{matrix}\right,$$

Rezolvați pentru K

$$K=-\frac{2q^{N}}{3}+12$$

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

Rezolvați pentru N

$$\left\{\begin{matrix}N=\log_{q}\left(-\frac{3K}{2}+18\right)\text{, }&K<12\text{ and }q\neq 1\text{ and }q>0\\N\in \mathrm{R}\text{, }&\left(q=1\text{ and }K=\frac{34}{3}\right)\text{ or }\left(q=-1\text{ and }K=\frac{38}{3}\text{ and }Denominator(N)\text{bmod}2=1\text{ and }Numerator(N)\text{bmod}2=1\right)\\N>0\text{, }&q=0\text{ and }K=12\end{matrix}\right,$$