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Zuletzt geändert von Holger Engels am 2025/03/13 07:51
Von Version 111.4
bearbeitet von Martina Wagner
am 2025/02/26 15:25
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Auf Version 68.1
bearbeitet von Martin Rathgeb
am 2025/02/25 21:07
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21 21  x*y = e --> y = e / x
22 22  e^y = x --> y = ln(x)
23 23  
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 jeweils eine passende Gleichung:
26 -
27 -Die Gleichung kann ich nach x auflösen, indem ich{{formula}} \ldots {{/formula}}
24 +{{aufgabe id="Exponentialgleichungen lösen (Fehlvorstellungen)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="5"}}
28 28  (% class="abc" %)
29 -1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
30 -1. {{formula}} \ldots {{/formula}} von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
31 -1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
26 +1. (((Beurteile folgende Aussagen:
27 +1) Die Gleichung {{formula}} 5^x = 2 {{/formula}} kann ich nach x auflösen, indem ich durch 5 dividiere. Ich erhalte damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}}.
28 +2) Die Gleichung {{formula}} 5^x = 2 {{/formula}} kann ich nach x auflösen, indem ich die 5-te Wurzel verwende. Ich erhalte damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}}.
29 +3) Um die Gleichung {{formula}} 5^x = 2 {{/formula}} nach x aufzulösen, benötige ich eine neue Methode bzw. eine neue Operation.
30 +)))
31 +1. Umkehraufgabe: Gib für jede in a) falsche Methode eine passende Gleichung an.
32 32  {{/aufgabe}}
33 33  
34 -{{aufgabe id="Gleichungen aufstellen II" afb="I" kompetenzen="K2,K5" quelle="Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
34 +{{aufgabe id="Gleichungsformen instantiieren" afb="I" kompetenzen="K2,K5" quelle="Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
35 35  Nenne möglichst viele (wahre) Gleichungen der folgenden Formen, wobei {{formula}} a, b, c \in \{2; 3; 4; \ldots; 16\} {{/formula}} gelten soll:
36 -{{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:; \qquad c = a\cdot b\:. {{/formula}}
36 +{{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:. {{/formula}}
37 37  {{/aufgabe}}
38 38  
39 39  {{aufgabe id="Darstellungen zuordnen" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
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}}|(((
44 -|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 -|y|0|1|8|27
50 -)))|[[image:x^3und8.svg||width="200px"]]
51 -
40 +Ordne zu!
52 52  (% class="abc" %)
53 -1. (((Gleichungen (implizite und explizite):
54 -1. {{formula}} x^3 = 8 {{/formula}}
55 -1. {{formula}} 2^x = 8 {{/formula}}
56 -1. {{formula}} x = \sqrt[3]{8=} {{/formula}}
57 -1. {{formula}} x = \log_{2}(8) {{/formula}}
58 -)))
59 -1. Wertetabellen:
60 -(((
61 -|x|0|1|2|3
62 -|y|0|1|8|27
63 -)))
64 -
65 -(((
66 -|x|0|1|2|3
67 -|y|0|1|8|27
68 -)))
42 +1. vier Gleichungen
43 +1. zwei Tabellen
69 69  1. zwei Graphen
70 -[[image:2^xund8.svg||width="200px"]]
71 -[[image:x^3und8.svg||width="200px"]]
72 72  {{/aufgabe}}
73 73  
74 74  {{aufgabe id="Logarithmen auswerten" afb="II" kompetenzen="K4,K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
75 -Ordne (ohne WTR!) die Terme ihren Werten gemäß den Kästchen über dem Zahlenstrahl zu. Trage dafür die jeweiligen Buchstaben in die Kästchen ein.
48 +Ordne (ohne WTR) die Terme ihren Werten gemäß den Kästchen über dem Zahlenstrahl zu. Trage dafür die jeweiligen Buchstaben in die Kästchen ein.
76 76  
77 -[[image:Logarithmus_neu.svg||width="600px"]]
50 +[[image:Logarithmus.svg||width="600px"]]
78 78  
79 79  (% class="abc" %)
80 -1. {{formula}} \log_{10}(0.1) {{/formula}}
81 -1. {{formula}} \log_{100}(0.1) {{/formula}}
82 -1. {{formula}} \log_{0.1}(0.1) {{/formula}}
53 +1. {{formula}} \log_{10}(10) {{/formula}}
54 +1. {{formula}} \log_{100}(10) {{/formula}}
55 +1. {{formula}} \log_{11}(10) {{/formula}}
83 83  1. {{formula}} \log_{10}(1000) {{/formula}}
84 -1. {{formula}} \log_{10}(50) {{/formula}}
85 -1. {{formula}} \log_{0.1}(1000) {{/formula}}
57 +1. {{formula}} \log_{10}(5) {{/formula}}
58 +1. {{formula}} \log_{11}(1000) {{/formula}}
86 86  1. {{formula}} \log_{10}(1) {{/formula}}
87 87  1. {{formula}} \log_{100}(10) {{/formula}}
88 88  1. {{formula}} \log_{10}(10) {{/formula}}
89 89  {{/aufgabe}}
90 90  
91 -{{aufgabe id="Exponentialgleichungen lösen (graphisch versus rechnerisch)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
64 +{{aufgabe id="Exponentialgleichungen lösen (graphisch vs rechnerisch)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
92 92  (% class="abc" %)
93 93  Ermittle die Lösung der Gleichung {{formula}} 2^x = 5 {{/formula}} graphisch und rechnerisch.
94 94  {{/aufgabe}}
95 95  
96 -{{aufgabe id="Exponentialgleichungen Lösbarkeit (graphisch versus rechnerisch)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
69 +{{aufgabe id="Exponentialgleichungen Lösbarkeit (graphisch vs rechnerisch)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
97 97  (% class="abc" %)
98 -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}}.
99 -{{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:. {{/formula}}
71 +Gegeben sind die beiden Gleichungen {{formula}} 2^x = y_0 \qquad x^2 = y_0 {{/formula}}. Untersuche ihre Lösbarkeit in Abhängigkeit von {{formula}} y_0 {{/formula}}.
100 100  {{/aufgabe}}
101 101  
102 -
103 -{{aufgabe id="Gleichungstypen einstudieren" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="20"}}
104 -Bestimme die Lösung der folgenden Gleichungen:
105 -
106 -(% class="border slim " %)
107 -|Typ 1 Umkehroperationen|Typ 2 Ausklammern|Typ 3 Substitution
108 -|{{formula}}x^2 = 2{{/formula}}|{{formula}}x^2-2x = 0{{/formula}}|{{formula}}x^4-40x^2+144 = 0{{/formula}}
109 -|{{formula}}x^4 = e{{/formula}}|{{formula}}2x^e = x^{2e}{{/formula}}|{{formula}}x^{2x}+x^e+1 = 0{{/formula}}
110 -|{{formula}}e^x = e{{/formula}}|{{formula}}2e^x = e^{2x}{{/formula}}|{{formula}}10^{6x}-2\cdot 10^{3x}+1 = 0{{/formula}}
111 -|{{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}}
112 -{{/aufgabe}}
113 -
114 -Nenne eine passende Gleichung. Die Gleichung kann ich nach x auflösen, indem ich {{formula}} \ldots {{/formula}}
115 -(% class="abc" %)
116 -1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
117 -1. {{formula}} \ldots {{/formula}} von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
118 -1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
119 -{{/aufgabe}}
120 -
121 -
122 122  {{aufgabe id="Exponentialgleichungen (Logarithmieren)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="15"}}
123 123  Bestimme die Lösungsmenge der Exponentialgleichung:
124 124  (% class="abc" %)
... ... @@ -129,7 +129,7 @@
129 129  1. {{formula}} e^x=-5 {{/formula}}
130 130  {{/aufgabe}}
131 131  
132 -{{aufgabe id="Exponentialgleichungen (Nullproduktsatz)" afb="II" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="12"}}
84 +{{aufgabe id="Exponentialgleichungen (Ausklammern, SVNP)" afb="II" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="12"}}
133 133  Bestimme die Lösungsmenge der Gleichung:
134 134  (% class="abc" %)
135 135  1. {{formula}} 2x=x^{2} {{/formula}}
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158 158  {{aufgabe id="Exponentialgleichungen graphisch" afb="II" kompetenzen="K4,K6" quelle="Niklas Wunder" cc="BY-SA" zeit="5"}}
159 159  Löse mit Hilfe der nebenstehenden Abbildung folgende Exponentialgleichungen näherungsweise. Hinweis: Ordne die linke und die rechte Seite der jeweiligen Gleichung passend den Funktionsgraphen zu.
160 160  (% class="abc" %)
161 -1. {{formula}} 2^x=(\frac{3}{4})^x+2 {{/formula}}
162 -1. {{formula}} 7-e^{x-3}=(\frac{3}{4})^x+2 {{/formula}}
163 -1. {{formula}} 2^x=1{,}5^{x+2}-0{,}5 {{/formula}}
164 -1. {{formula}} 7-e^{x-3}=4-\frac{1}{2}\,x {{/formula}}
113 +a) {{formula}} 2^x=(\frac{3}{4})^x+2 {{/formula}}
114 +b) {{formula}} 7-e^{x-3}=(\frac{3}{4})^x+2 {{/formula}}
115 +c) {{formula}} 2^x=1{,}5^{x+2}-0{,}5 {{/formula}}
116 +d) {{formula}} 7-e^{x-3}=4-\frac{1}{2}\,x {{/formula}}
165 165  
166 -[[image:ExpGlei.svg||width="600px"]]
118 +[[image:ExpGlei.svg]]
167 167  {{/aufgabe}}
168 168  
169 169  {{seitenreflexion/}}
2^xund8.ggb
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42 -
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x^3 und 8.svg
Author
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1 +XWiki.martinrathgeb
Größe
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1 +6.7 KB
Inhalt
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