$N(x)=N_0\cdot e^{k\cdot (x-2000)}N(x)=N\_0\backslash cdot\; e^\{k\backslash cdot\; (x-2000)\}$

Setze die Werte für das Jahr $20002000$ und $20102010$ ein.

$N(2000)=\mathrm{6,1}\text{Mrd}=N_0\cdot e^{k\cdot (2000-2000)}N(2000)=6\{,\}1\backslash ;\; \backslash text\{Mrd\}\backslash ;=N\_0\backslash cdot\; e^\{k\backslash cdot(2000-2000)\}$

$N(2010)=\mathrm{6,9}\text{Mrd}=N_0\cdot e^{k\cdot (2010-2000)}N(2010)=6\{,\}9\; \backslash ;\; \backslash text\{Mrd\}\backslash ;=N\_0\backslash cdot\; e^\{k\backslash cdot(2010-2000)\}$

$N(2000)=\mathrm{6,1}\text{Mrd}=N_0\cdot e^{k\cdot (0)}N(2000)=6\{,\}1\backslash ;\; \backslash text\{Mrd\}\backslash ;=N\_0\backslash cdot\; e^\{k\backslash cdot(0)\}$

Vereinfache diese Gleichung.

$N(2000)=\mathrm{6,1}\text{Mrd}=N_0\cdot 1N(2000)=6\{,\}1\backslash ;\; \backslash text\{Mrd\}\backslash ;=N\_0\backslash cdot1$

$\mathrm{1,131}=e^{k\cdot 10}1\{,\}131=e^\{k\backslash cdot10\}$

Wende den Logarithmus an.

$\ln \left(\mathrm{1,131}\right)=10k\mathrm{\mid }:10\backslash ln\backslash left(1\{,\}131\backslash right)=10k\backslash ,\backslash ,|:10$

$k=\mathrm{0,0123}k=0\{,\}0123$