Réitigh 0=1+x_{H}+z^xH

$$0 = 1 + x _ { H } + z ^ { x } H$$

Réitigh do H. (complex solution)

$$\left\{\begin{matrix}H=-\frac{x_{H}+1}{z^{x}}\text{, }&x=0\text{ or }z\neq 0\\H\in \mathrm{C}\text{, }&x_{H}=-1\text{ and }z=0\text{ and }x\neq 0\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$1+x_{H}+z^{x}H=0$$

Bain 1 ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$x_{H}+z^{x}H=-1$$

Bain x_{H} ón dá thaobh.

$$z^{x}H=-1-x_{H}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$z^{x}H=-x_{H}-1$$

Roinn an dá thaobh faoi z^{x}.

$$\frac{z^{x}H}{z^{x}}=\frac{-x_{H}-1}{z^{x}}$$

Má roinntear é faoi z^{x} cuirtear an iolrúchán faoi z^{x} ar ceal.

$$H=\frac{-x_{H}-1}{z^{x}}$$

Roinn -1-x_{H} faoi z^{x}.

$$H=-\frac{x_{H}+1}{z^{x}}$$

Réitigh do H.

$$\left\{\begin{matrix}H=-\frac{x_{H}+1}{z^{x}}\text{, }&z>0\text{ or }\left(Denominator(x)\text{bmod}2=1\text{ and }z<0\right)\\H\in \mathrm{R}\text{, }&x_{H}=-1\text{ and }z=0\text{ and }x>0\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$1+x_{H}+z^{x}H=0$$

Bain 1 ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$x_{H}+z^{x}H=-1$$

Bain x_{H} ón dá thaobh.

$$z^{x}H=-1-x_{H}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$z^{x}H=-x_{H}-1$$

Roinn an dá thaobh faoi z^{x}.

$$\frac{z^{x}H}{z^{x}}=\frac{-x_{H}-1}{z^{x}}$$

Má roinntear é faoi z^{x} cuirtear an iolrúchán faoi z^{x} ar ceal.

$$H=\frac{-x_{H}-1}{z^{x}}$$

Roinn -1-x_{H} faoi z^{x}.

$$H=-\frac{x_{H}+1}{z^{x}}$$

Réitigh do x. (complex solution)

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

Réitigh do x.

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