Änderungen von Dokument Lösung Basiswechel

Zusammenfassung

• Seiteneigenschaften (2 geändert, 0 hinzugefügt, 0 gelöscht)

Details

Seiteneigenschaften

Dokument-Autor

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1		-XWiki.niklaswunder
	1	+XWiki.holgerengels

Inhalt

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1		-a) {{formula}}(\frac{1}{4})^x=(2^{-x})^x=2^{-2x}{{/formula}}
2		-b) {{formula}}(\frac{3}{18})^x=(\frac{1}{6})^=6^{-x}{{/formula}}
3		-c) {{formula}}5^{2x+1}=5\cdot (5^2)^x=5\cdot 25^x{{/formula}}
4		-d) {{formula}}(\frac{16}{54})^{2x}=(\frac{8}{27})^{2x}=(\frac{2}{3})^{3\cdot 2x}=(-\frac{2}{3})^{-6x}{{/formula}}
	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	+)))
	11	+