Wiki-Quellcode von Lösung Basiswechsel
Zuletzt geändert von akukin am 2025/08/11 15:24
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| author | version | line-number | content |
|---|---|---|---|
 |
1.1 | 1 | (% class="abc" %) |
| 2 | 1. ((({{formula}}f(x)=(\frac{1}{4})^x{{/formula}}, neue Basis {{formula}}b=2{{/formula}} | ||
| 3 | |||
| 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}} | ||
| 7 | |||
| |
1.2 | 8 | {{formula}}9 = \left(\frac{1}{3}\right)^{-2} \Rightarrow f(x)=\left(\frac{1}{3}\right)^{-2x}{{/formula}} |
| |
1.1 | 9 | ))) |
| 10 | 1. ((({{formula}}f(x)=5^{2x+1}{{/formula}}, neue Basis {{formula}}b=25{{/formula}} | ||
| 11 | |||
| 12 | {{formula}}5 = 25^{\frac{1}{2}} \Rightarrow f(x)=25^{\frac{1}{2}\cdot(2x+1)}=25^{x+\frac{1}{2}}{{/formula}} | ||
| 13 | ))) |