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zu a)$$2100\cdot\left(1+\frac{3,5}{100}\right)^7=\underline{2671,79}$$zu b)$$\left.2100\cdot\left(1+\frac{p}{100}\right)^{14}=4200\quad\right|\;:2100$$$$\left.\left(1+\frac{p}{100}\right)^{14}=2=\left(\sqrt[14]{2}\right)^{14}\quad\right|\;\sqrt[14]{\cdots}$$$$\left.1+\frac{p}{100}=\sqrt[14]{2}\quad\right|\;-1$$$$\left.\frac{p}{100}=\sqrt[14]{2}-1\quad\right|\;\cdot100$$$$\left.p=100\left(\sqrt[14]{2}-1\right)\quad\right.$$$$\underline{p=5,0757\%}$$zu c)$$\left.1900\cdot\left(1+\frac{4,5}{100}\right)^n>2100\cdot\left(1+\frac{3,5}{100}\right)^n\quad\right|\;:1900$$$$\left.\left(1+\frac{4,5}{100}\right)^n>\frac{2100}{1900}\cdot\left(1+\frac{3,5}{100}\right)^n\quad\right|\;\ln(\cdots)$$$$\left.n\ln\left(1+\frac{4,5}{100}\right)>\ln\left(\frac{2100}{1900}\right)+n\ln\left(1+\frac{3,5}{100}\right)\quad\right|\;-n\ln\left(1+\frac{3,5}{100}\right)$$$$\left.n\ln\left(1+\frac{4,5}{100}\right)-n\ln\left(1+\frac{3,5}{100}\right)>\ln\left(\frac{2100}{1900}\right)\quad\right|\;\text{ausklammern}$$$$\left.n\left(\ln\left(1+\frac{4,5}{100}\right)-\ln\left(1+\frac{3,5}{100}\right)\right)>\ln\left(\frac{2100}{1900}\right)\quad\right|\;\ln(\cdots)\text{ berechnen}$$$$\left.n\cdot0,00961546>0,1000835\quad\right|\;:0,00961546$$$$\underline{n>10,41}$$
