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$ B_{t} = B_{0} \cdot (1 - \frac{ p}{100})^{t} $
$ B_{0} = \frac{B_{t} }{(1 - \frac{ p}{100})^{t} } $
$ t =\frac{\ln(B_{t} ) - \ln(B_{0} )}{ \ln(1 - \frac{ p}{100})} $
$ p = (1 - ^{t} \sqrt{\frac{ B_{t} }{B_{0} }})\cdot 100 $
Algebra-Finanzmathematik-Degressive Abschreibung
$B_{t} = B_{0} \cdot (1 - \frac{ p}{100})^{t}$
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$B_{0} = \frac{B_{t} }{(1 - \frac{ p}{100})^{t} }$
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$t =\frac{\ln(B_{t} ) - \ln(B_{0} )}{ \ln(1 - \frac{ p}{100})}$
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$p = (1 - ^{t} \sqrt{\frac{ B_{t} }{B_{0} }})\cdot 100$
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Beispiel Nr: 04
$\begin{array}{l} \text{Gegeben:}\\\text{Anzahl der Jahre} \qquad t \qquad \\
\text{Abschreibungssatz} \qquad p \qquad [\%] \\
\text{Buchwert} \qquad B_{t} \qquad [Euro] \\
\\ \text{Gesucht:} \\\text{Anschaffungswert} \qquad B_{0} \qquad [Euro] \\
\\ B_{0} = \frac{B_{t} }{(1 - \frac{ p}{100})^{t} }\\ \textbf{Gegeben:} \\ t=1\frac{1}{14} \qquad p=1\frac{1}{18}\% \qquad B_{t}=5Euro \qquad \\ \\ \textbf{Rechnung:} \\B_{0} = \frac{B_{t} }{(1 - \frac{ p}{100})^{t} } \\
t=1\frac{1}{14}\\
p=1\frac{1}{18}\%\\
B_{t}=5Euro\\
B_{0} = \frac{ 5Euro }{(1 - \frac{1\frac{1}{18}\%}{100})^{1\frac{1}{14}} }\\\\B_{0}=5,06Euro
\\\\ \end{array}$