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Zuletzt geändert von Holger Engels am 2025/03/13 07:51
Von Version 92.1
bearbeitet von Dirk Tebbe
am 2025/02/26 14:15
Änderungskommentar: Löschung des Anhangs 8undx^3.ggb
Auf Version 113.1
bearbeitet von Elke Hallmann
am 2025/02/26 15:31
Änderungskommentar: Es gibt keinen Kommentar für diese Version

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1 +XWiki.elkehallmanngmxde
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22 22  e^y = x --> y = ln(x)
23 23  
24 24  {{aufgabe id="Gleichungen aufstellen I" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="5"}}
25 -Nenne eine passende Gleichung. Die Gleichung kann ich nach x auflösen, indem ich {{formula}} \ldots {{/formula}}
25 +Nenne jeweils eine passende Gleichung:
26 +
27 +Die Gleichung kann ich nach x auflösen, indem ich{{formula}} \ldots {{/formula}}
26 26  (% class="abc" %)
27 27  1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
28 28  1. {{formula}} \ldots {{/formula}} von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
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35 35  {{/aufgabe}}
36 36  
37 37  {{aufgabe id="Darstellungen zuordnen" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
38 -Ordne zu!
39 -(% class="abc" %)
40 -1. (((Gleichungen (implizite und explizite):
41 -1. {{formula}} x^3 = 8 {{/formula}}
42 -1. {{formula}} 2^x = 8 {{/formula}}
43 -1. {{formula}} x = \sqrt[3]{8=} {{/formula}}
44 -1. {{formula}} x = \log_{2}(8) {{/formula}}
45 -)))
46 -1. Wertetabellen:
47 -(((
40 +Ordne zu:
41 +(% class="border slim " %)
42 +|Implizite Gleichungen|Explizite Gleichungen|Wertetabellen|Schaubilder
43 +|{{formula}} x^{-3} = 8 {{/formula}}|{{formula}} x = \sqrt[3]{8} {{/formula}}|(((
48 48  |x|0|1|2|3
45 +|y|1|2|4|8
46 +)))|[[image:2^xund8.svg||width="200px"]]
47 +|{{formula}} 2^x = 8 {{/formula}}|{{formula}} x = -\log_{2}(8) {{/formula}} |(((
48 +|x|0|1|2|3
49 49  |y|0|1|8|27
50 -)))
51 -
52 -(((
50 +)))|[[image:x^3und8.svg||width="200px"]]
51 +|{{formula}} 2^{-x} = 8 {{/formula}}|{{formula}} x = \log_{2}(8) {{/formula}} |(((
53 53  |x|0|1|2|3
54 54  |y|0|1|8|27
55 -)))
56 -1. zwei Graphen
57 -[[image:8und2^x.svg||width="200px"]]
58 -[[image:x^3 und 8.svg||width="200px"]]
54 +)))|[[image:x^3und8.svg||width="200px"]]
55 +|{{formula}} 2^x = 8 {{/formula}}|{{formula}} x = x = \frac{1}{\sqrt[3]{8}} {{/formula}} |(((
56 +|x|0|1|2|3
57 +|y|0|1|8|27
58 +)))|[[image:x^3und8.svg||width="200px"]]
59 +
60 +
59 59  {{/aufgabe}}
60 60  
61 61  {{aufgabe id="Logarithmen auswerten" afb="II" kompetenzen="K4,K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
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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}}
69 +1. {{formula}} \log_{10}(0.1) {{/formula}}
70 +1. {{formula}} \log_{100}(0.1) {{/formula}}
71 +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}}
74 +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}}
76 76  {{/aufgabe}}
77 77  
78 -{{aufgabe id="Exponentialgleichungen lösen (graphisch vs rechnerisch)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
80 +{{aufgabe id="Exponentialgleichungen lösen (graphisch versus rechnerisch)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
79 79  (% class="abc" %)
80 80  Ermittle die Lösung der Gleichung {{formula}} 2^x = 5 {{/formula}} graphisch und rechnerisch.
81 81  {{/aufgabe}}
82 82  
83 -{{aufgabe id="Exponentialgleichungen Lösbarkeit (graphisch vs rechnerisch)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
85 +{{aufgabe id="Exponentialgleichungen Lösbarkeit (graphisch versus rechnerisch)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
84 84  (% class="abc" %)
85 85  Gegeben sind die beiden Gleichungen {{formula}} x^2 = a {{/formula}} und {{formula}} 2^x = a {{/formula}} für {{formula}} a \in \mathbb{R} {{/formula}}. Untersuche ihre Lösbarkeit in Abhängigkeit von {{formula}} a {{/formula}}.
86 86  {{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:. {{/formula}}
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87 87  {{/aufgabe}}
88 88  
89 89  
90 -{{aufgabe id="Gleichungstypen einstudieren" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="5"}}
92 +{{aufgabe id="Gleichungstypen einstudieren" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="20"}}
91 91  Bestimme die Lösung der folgenden Gleichungen:
92 92  
93 93  (% class="border slim " %)
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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
99 +|{{formula}}e^x = e{{/formula}}|{{formula}}2e^x = e^{2x}{{/formula}}|{{formula}}10^{6x}-2\cdot 10^{3x}+1 = 0{{/formula}}
100 +|{{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}}
2^x und 8.svg
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38 -
8und2^x.svg
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8undx^3.svg
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x^3 und 8.svg
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