\(2)\quad V_{0} = 1,2, a = 100% + 170% = 2,7\)
\(\Rightarrow V(t) = V_{0} * a^t = 1,2 * 2,7^t\)

\(V(t) = V_0 * a^t = V_0 * e^{\lambda * t}\)
\(\Leftrightarrow e^{\lambda * t} = a \Leftrightarrow \lambda * t = t * ln(a) \)
\(\Leftrightarrow \lambda = \frac{t * ln(a)}{t} = ln(a) = ln(2,7)\)

\(\Rightarrow V(t) = V_{0} * e^{\lambda * t} = 1,2 * e^{ln(2,7) * t}\)

\(3)\quad N_0 = 140, Sei\;a = 2 \) (vereinfacht Rechnung, da Verdopplungszeit gegeben)
\(N(14) = N_0 * 2^{\lambda * 10} \land N(14) = N_0 * 2 \Rightarrow \lambda * 14 = 1\)
\(\Leftrightarrow \lambda = \frac{1}{14} \Rightarrow N(t) = N_0 * a^{\lambda * t} = 140 * 2^{\frac{t}{14}}\)

Staatsverschuldung 2007: \(N(7) = 140 * 2^{\frac{7}{14}} =140 * \sqrt{2} \approx 198 \)
Staatsverschuldung 2030: \(N(30) = 140 * 2^{\frac{30}{14}} \approx 618 \)

\(4)\quad N_0:\) Atome zu Beginn \(, N_0 = 10^9, a = \frac{1}{2}^{\frac{1}{3,824}}\)
\(\Rightarrow^{aus Angabe} \lambda = -\ln{a} \approx -0,1813\)
\(\Rightarrow N(t) = N_0 * e^{\lambda * t} = 10^9 * e^{-0,1813t}\)

\(5)\quad \lambda = -0,0125\)
\(e^{\lambda * t_{\frac{1}{2}}} = \frac{1}{2} \Leftrightarrow \lambda * t_{\frac{1}{2}} = \ln{\frac{1}{2}}\Leftrightarrow t_{\frac{1}{2}} = \frac{\ln{\frac{1}{2}}}{\lambda}\)
\(\Rightarrow t_{\frac{1}{2}} = \frac{\ln{\frac{1}{2}}}{-0,0125}\approx 55,45\)