Wiki-Quellcode von Lösung Basiswechel
Zuletzt geändert von Holger Engels am 2025/03/10 21:46
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
2.1 | 1 | (% class="abc" %) |
| 2 | 1. ((({{formula}}f(x)=(\frac{1}{4})^x{{/formula}}, neue Basis {{formula}}b=2{{/formula}} | ||
| |
2.2 | 3 | |
| |
2.1 | 4 | {{formula}}\frac{1}{4} = 2^{-2} \Rightarrow f(x)=2^{-2x}{{/formula}} |
| 5 | ))) | ||
| 6 | 1. ((({{formula}}f(x)=9^x{{/formula}}, neue Basis {{formula}}b=\frac{1}{3}{{/formula}} | ||
| |
2.2 | 7 | |
| |
2.1 | 8 | {{formula}}9 = (\frac{1}{3})^{-2} \Rightarrow f(x)=(\frac{1}{3})^{-2x}{{/formula}} |
| 9 | ))) | ||
| 10 | 1. ((({{formula}}f(x)=5^{2x+1}{{/formula}}, neue Basis {{formula}}b=25{{/formula}} | ||
| |
2.2 | 11 | |
| |
2.1 | 12 | {{formula}}5 = 25^{\frac{1}{2}} \Rightarrow f(x)=25^{\frac{1}{2}\cdot(2x+1)}=25^{x+\frac{1}{2}}{{/formula}} |
| 13 | ))) | ||
| 14 |