Änderungen von Dokument BPE 4.5 Logarithmus und Exponentialgleichungen

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...	...	@@ -64,12 +64,12 @@
64	64	[[image:Logarithmus_neu.svg||width="600px"]]
65	65
66	66	(% class="abc" %)
67		-1. {{formula}} \log_{10}(10) {{/formula}}
68		-1. {{formula}} \log_{100}(10) {{/formula}}
69		-1. {{formula}} \log_{11}(10) {{/formula}}
	67	+1. {{formula}} \log_{10}(0.1) {{/formula}}
	68	+1. {{formula}} \log_{100}(0.1) {{/formula}}
	69	+1. {{formula}} \log_{0.1}(0.1) {{/formula}}
70	70	1. {{formula}} \log_{10}(1000) {{/formula}}
71	71	1. {{formula}} \log_{10}(50) {{/formula}}
72		-1. {{formula}} \log_{11}(1000) {{/formula}}
	72	+1. {{formula}} \log_{0.1}(1000) {{/formula}}
73	73	1. {{formula}} \log_{10}(1) {{/formula}}
74	74	1. {{formula}} \log_{100}(10) {{/formula}}
75	75	1. {{formula}} \log_{10}(10) {{/formula}}
...	...	@@ -94,8 +94,8 @@
94	94	|Typ 1 Umkehroperationen|Typ 2 Ausklammern|Typ 3 Substitution
95	95	|{{formula}}x^2 = 2{{/formula}}|{{formula}}x^2-2x = 0{{/formula}}|{{formula}}x^4-40x^2+144 = 0{{/formula}}
96	96	|{{formula}}x^4 = e{{/formula}}|{{formula}}2x^e = x^{2e}{{/formula}}|{{formula}}x^{2x}+x^e+1 = 0{{/formula}}
97		-|{{formula}}e^x = e{{/formula}}|{{formula}}2e^x = e^{2x}{{/formula}}|2
98		-|{{formula}}f_4(x){{/formula}}|2|1
	97	+|{{formula}}e^x = e{{/formula}}|{{formula}}2e^x = e^{2x}{{/formula}}|{{formula}}10^{6x}-2\cdot 10^{3x}+1 = 0{{/formula}}
	98	+|{{formula}}3e^x = \frac{1}{2}e^{-x}{{/formula}}|{{formula}}x\cdot 3^x+4\cdot 3^x = 0{{/formula}}|{{formula}}3e^x-1 = \frac{1}{3}e^{-x}{{/formula}}
99	99	{{/aufgabe}}
100	100
101	101	Nenne eine passende Gleichung. Die Gleichung kann ich nach x auflösen, indem ich {{formula}} \ldots {{/formula}}