Wiki-Quellcode von Lösung Basiswechel

Version 2.1 von Holger Engels am 2025/03/10 21:44

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1 (% class="abc" %)
2 1. ((({{formula}}f(x)=(\frac{1}{4})^x{{/formula}}, neue Basis {{formula}}b=2{{/formula}}
3 {{formula}}\frac{1}{4} = 2^{-2} \Rightarrow f(x)=2^{-2x}{{/formula}}
4 )))
5 1. ((({{formula}}f(x)=9^x{{/formula}}, neue Basis {{formula}}b=\frac{1}{3}{{/formula}}
6 {{formula}}9 = (\frac{1}{3})^{-2} \Rightarrow f(x)=(\frac{1}{3})^{-2x}{{/formula}}
7 )))
8 1. ((({{formula}}f(x)=5^{2x+1}{{/formula}}, neue Basis {{formula}}b=25{{/formula}}
9 {{formula}}5 = 25^{\frac{1}{2}} \Rightarrow f(x)=25^{\frac{1}{2}\cdot(2x+1)}=25^{x+\frac{1}{2}}{{/formula}}
10 )))