Resolver A=P(1+R/100)^n

$$A = P ( 1 + \frac { R } { 100 } ) ^ { n }$$

Ebatzi: P (complex solution)

$$\left\{\begin{matrix}P=\frac{A}{\left(\frac{R+100}{100}\right)^{n}}\text{, }&n=0\text{ or }R\neq -100\\P\in \mathrm{C}\text{, }&A=0\text{ and }R=-100\text{ and }n\neq 0\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$P\left(1+\frac{R}{100}\right)^{n}=A$$

Modu arruntean dago ekuazioa.

$$\left(\frac{R}{100}+1\right)^{n}P=A$$

Zatitu ekuazioaren bi aldeak \left(1+\frac{1}{100}R\right)^{n} balioarekin.

$$\frac{\left(\frac{R}{100}+1\right)^{n}P}{\left(\frac{R}{100}+1\right)^{n}}=\frac{A}{\left(\frac{R}{100}+1\right)^{n}}$$

\left(1+\frac{1}{100}R\right)^{n} balioarekin zatituz gero, \left(1+\frac{1}{100}R\right)^{n} balioarekiko biderketa desegiten da.

$$P=\frac{A}{\left(\frac{R}{100}+1\right)^{n}}$$

Ebatzi: P

$$\left\{\begin{matrix}P=\frac{A}{\left(\frac{R+100}{100}\right)^{n}}\text{, }&R>-100\text{ or }\left(Denominator(n)\text{bmod}2=1\text{ and }R<-100\right)\\P\in \mathrm{R}\text{, }&A=0\text{ and }R=-100\text{ and }n>0\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$P\left(1+\frac{R}{100}\right)^{n}=A$$

Modu arruntean dago ekuazioa.

$$\left(\frac{R}{100}+1\right)^{n}P=A$$

Zatitu ekuazioaren bi aldeak \left(1+\frac{1}{100}R\right)^{n} balioarekin.

$$\frac{\left(\frac{R}{100}+1\right)^{n}P}{\left(\frac{R}{100}+1\right)^{n}}=\frac{A}{\left(\frac{R}{100}+1\right)^{n}}$$

\left(1+\frac{1}{100}R\right)^{n} balioarekin zatituz gero, \left(1+\frac{1}{100}R\right)^{n} balioarekiko biderketa desegiten da.

$$P=\frac{A}{\left(\frac{R}{100}+1\right)^{n}}$$

Ebatzi: A (complex solution)

$$A=P\times \left(\frac{R+100}{100}\right)^{n}$$

Ebatzi: A

$$A=P\times \left(\frac{R+100}{100}\right)^{n}$$

$$\left(R<-100\text{ and }Denominator(n)\text{bmod}2=1\right)\text{ or }\left(R=-100\text{ and }n>0\right)\text{ or }R>-100$$