Die Formel für den exponentiellen Zerfall lautet:

$N\left(t\right)=N_0\cdot (1-p{)}^t\backslash hspace\{4cm\}N\backslash left(t\backslash right)=N\_0\backslash cdot(1-p)^t$

$N_0=50000p_1=9\mathrm{\%}{t}_1=2\text{ Jahre};p_2=8\mathrm{\%}{t}_2=4\text{ Jahre}N\_0\backslash \; =\backslash \; 50000\backslash qquad\; p\_\{1\}\backslash \; =\backslash \; 9\backslash \%\backslash \; \backslash \; t\_\{1\}\backslash \; =\backslash \; 2\backslash \; \backslash text\{Jahre\};\backslash qquad\; p\_\{2\}\backslash \; =\backslash \; 8\backslash \%\backslash \; \backslash \; t\_\{2\}\backslash \; =\backslash \; 4\backslash \; \backslash text\{Jahre\}\backslash qquad$

Berechne den Wert des Wohnmobils nach 6 Jahren

$N\left(6\right)=50000\cdot (1-\mathrm{0,09}{)}^2\cdot (1-\mathrm{0,08}{)}^4\backslash \; N\backslash left(6\backslash right)=50000\backslash cdot(1-0\{,\}09)^2\backslash cdot(1-0\{,\}08)^4$

$N\left(6\right)=50000\cdot {\mathrm{0,91}}^2\cdot {\mathrm{0,92}}^4\backslash \; N\backslash left(6\backslash right)=50000\backslash cdot0\{,\}91^2\backslash cdot0\{,\}92^4$

$N\left(6\right)=50000\cdot \mathrm{0,593245}\underset{‾}{\underset{‾}{N\left(6\right)\approx 29662\text{ €}}}\backslash \; N\backslash left(6\backslash right)=50000\backslash cdot0\{,\}593245\backslash \backslash \backslash \; \backslash underline\{\backslash underline\{N\backslash left(6\backslash right)\backslash approx29662\backslash \; €\}\}$

Nach $66$ Jahren beträgt der Wert des Wohnmobils noch ungefähr $2966229662$ Euro.

$N\left(t\right)=N_0\cdot (1-p{)}^t\backslash hspace\{4cm\}N\backslash left(t\backslash right)=N\_0\backslash cdot(1-p)^t$

$N_t=19500N_0=45000t=12\text{ Jahre}N\_t\backslash \; =\backslash \; 19500\backslash qquad\; N\_\{0\}\backslash \; =\backslash \; 45000\backslash qquad\; t\backslash \; =\backslash \; 12\backslash \; \backslash text\{Jahre\}\backslash qquad$setze die Werte in die Formel ein.

$19500=45000\cdot (1-p{)}^{12}\backslash hspace\{4cm\}19500=45000\backslash cdot(1-p)^\{12\}$

$19500=45000\cdot (1-p{)}^{12}\Rightarrow (1-p)=\sqrt[12]{{\displaystyle \frac{19500}{45000}}}\Rightarrow (1-p)\approx \mathrm{0,933}\underset{‾}{\underset{‾}{p\approx \mathrm{6,7}\mathrm{\%}}}19500\backslash \; =\backslash \; 45000\backslash \; \backslash cdot\backslash \; (1-p)^\{12\}\backslash \; \backslash Rightarrow\backslash \; (1-p)\backslash \; =\backslash \; \backslash sqrt[\backslash scriptsize\{12\}]\{\backslash dfrac\{19500\}\{45000\}\}\backslash quad\backslash Rightarrow\backslash quad(1-p)\backslash \; \backslash approx\backslash \; 0\{,\}933\backslash \; \backslash \backslash \backslash qquad\; \backslash underline\{\backslash underline\{p\backslash approx6\{,\}7\backslash \%\}\}$

Der Wertverlust pro Jahr beträgt ungefähr $\mathrm{6,7}\mathrm{\%}6\{,\}7\backslash \%$.

Die Formel für den exponentiellen Zerfall lautet:

$N\left(t\right)=N_0\cdot (1-p{)}^t\backslash hspace\{4cm\}N\backslash left(t\backslash right)=N\_0\backslash cdot(1-p)^t$

$N_t=9750\text{ €}p=\mathrm{6,6}\mathrm{\%}{N}_0=19500\text{ €}N\_t\backslash \; =\backslash \; 9750\backslash \; \backslash text\{€\}\backslash qquad\; p\backslash \; =\backslash \; 6\{,\}6\backslash \%\backslash qquad\; N\_\{0\}\backslash \; =\backslash \; 19500\backslash \; \backslash text\{€\}\backslash quad$setze die Werte in die Formel ein.

$9750=19500\cdot (1-\mathrm{0,066}{)}^t\backslash hspace\{4cm\}9750=19500\backslash cdot(1-0\{,\}066)^t$

$(1-\mathrm{0,066}{)}^t={\displaystyle \frac{9750}{19500}}\Rightarrow (\mathrm{0,934}{)}^t=\mathrm{0,5}\Rightarrow t={\displaystyle \frac{\text{ln }\mathrm{0,5}}{\text{ln }\mathrm{0,934}}}\backslash \; (1-0\{,\}066)^t=\backslash dfrac\{9750\}\{19500\}\backslash quad\backslash Rightarrow\backslash quad(0\{,\}934)^t=0\{,\}5\backslash quad\backslash Rightarrow\backslash quad\; t=\backslash dfrac\{\backslash text\{ln\}\backslash \; 0\{,\}5\}\{\backslash text\{ln\}\backslash \; 0\{,\}934\}$

$\underset{‾}{\underset{‾}{t\approx 10\text{ Jahre}}}\backslash hspace\{80mm\}\backslash underline\{\backslash underline\{t\backslash approx10\backslash \; \backslash text\{Jahre\}\}\}$

Die Familie Grün müsste ihr Wohnmobil nach ungefähr $1010$ Jahren verkaufen um noch mindestens die Hälfte des Kaufpreises von $19500\text{ €}19500\backslash \; €$ zu erhalten.