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Zuletzt geändert von Holger Engels am 2025/03/13 07:51
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Auf Version 87.1
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1 -XWiki.holgerengels
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7 7  [[Kompetenzen.K4]] [[Kompetenzen.K6]] Ich kann die Lösungen einer Exponentialgleichung als Nullstelle interpretieren
8 8  [[Kompetenzen.K4]] [[Kompetenzen.K6]] Ich kann die Lösungen einer Exponentialgleichung als Schnittstelle zweier Funktionen interpretieren
9 9  
10 -{{lehrende}}
11 11  Aufgaben:
12 12  – Logarithmus: graphisches Ermitteln vs. Operator
13 13  Lösen von Exponentialgleichungen:
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20 20  Gleichungen:
21 21  x+y = e --> y = e - x
22 22  x*y = e --> y = e / x
23 -e^y = x --> y = {{{ln(x)}}}
24 -{{/lehrende}}
22 +e^y = x --> y = ln(x)
25 25  
26 26  {{aufgabe id="Gleichungen aufstellen I" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="5"}}
27 -Nenne jeweils eine passende Gleichung:
28 -
29 -Die Gleichung kann ich nach x auflösen, indem ich …
25 +Nenne eine passende Gleichung. Die Gleichung kann ich nach x auflösen, indem ich {{formula}} \ldots {{/formula}}
30 30  (% class="abc" %)
31 -1. … die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
32 -1. … von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
33 -1. … die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
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 +1. {{formula}} \ldots {{/formula}} von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
29 +1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
34 34  {{/aufgabe}}
35 35  
36 36  {{aufgabe id="Gleichungen aufstellen II" afb="I" kompetenzen="K2,K5" quelle="Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
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38 38  {{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:; \qquad c = a\cdot b\:. {{/formula}}
39 39  {{/aufgabe}}
40 40  
41 -{{aufgabe id="Darstellungen zuordnen" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="6"}}
42 -Ordne zu:
43 -(% class="border slim " %)
44 -|Implizite Gleichungen|Explizite Gleichungen|Wertetabellen|Schaubilder
45 -|{{formula}} x^{-3} = 8 {{/formula}}|{{formula}} x = \sqrt[3]{8} {{/formula}}|(((
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 +(((
46 46  |x|0|1|2|3
47 -|y|1|2|4|8
48 -)))|[[image:2^xund8.svg||width="200px"]]
49 -|{{formula}} 2^x = 8 {{/formula}}|{{formula}} x = -\log_{2}(8) {{/formula}} |(((
50 -|x|0|1|2|3
51 51  |y|0|1|8|27
52 -)))|[[image:2^-xund8.svg||width="200px"]]
53 -|{{formula}} 2^{-x} = 8 {{/formula}}|{{formula}} x = \log_{2}(8) {{/formula}} |(((
50 +)))
51 +
52 +(((
54 54  |x|0|1|2|3
55 -|y|1|\frac{1}{2}|\frac{1}{4}|\frac{1}{8}
56 -)))|[[image:x^3und8.svg||width="200px"]]
57 -|{{formula}} 2^x = 8 {{/formula}}|{{formula}} x = x = \frac{1}{\sqrt[3]{8}} {{/formula}} |(((
58 -|x|0|1|2|3
59 -|y|n.d.|1|\frac{1}{8}|\frac{1}{27}
60 -)))|[[image:x^-3und8.svg||width="200px"]]
54 +|y|0|1|8|27
55 +)))
56 +1. zwei Graphen
57 +[[image:8und2^x.svg||width="200px"]]
61 61  {{/aufgabe}}
62 62  
63 63  {{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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66 66  [[image:Logarithmus_neu.svg||width="600px"]]
67 67  
68 68  (% class="abc" %)
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}}
66 +1. {{formula}} \log_{10}(10) {{/formula}}
67 +1. {{formula}} \log_{100}(10) {{/formula}}
68 +1. {{formula}} \log_{11}(10) {{/formula}}
72 72  1. {{formula}} \log_{10}(1000) {{/formula}}
73 73  1. {{formula}} \log_{10}(50) {{/formula}}
74 -1. {{formula}} \log_{0.1}(1000) {{/formula}}
71 +1. {{formula}} \log_{11}(1000) {{/formula}}
75 75  1. {{formula}} \log_{10}(1) {{/formula}}
76 76  1. {{formula}} \log_{100}(10) {{/formula}}
77 77  1. {{formula}} \log_{10}(10) {{/formula}}
78 78  {{/aufgabe}}
79 79  
80 -{{aufgabe id="Exponentialgleichungen lösen (graphisch versus rechnerisch)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
77 +{{aufgabe id="Exponentialgleichungen lösen (graphisch vs rechnerisch)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="5"}}
81 81  (% class="abc" %)
82 82  Ermittle die Lösung der Gleichung {{formula}} 2^x = 5 {{/formula}} graphisch und rechnerisch.
83 83  {{/aufgabe}}
84 84  
85 -{{aufgabe id="Gleichungen gemeinsamer Form" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
82 +{{aufgabe id="Exponentialgleichungen Lösbarkeit (graphisch vs rechnerisch)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
86 86  (% class="abc" %)
87 -Aufgabe als Dokument im Anhang ‚unten‘.
84 +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}}.
85 +{{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:. {{/formula}}
88 88  {{/aufgabe}}
89 89  
90 -{{aufgabe id="Gleichungstypen einstudieren" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="20"}}
91 -Bestimme die Lösung der folgenden Gleichungen:
92 92  
93 -(% class="border slim " %)
94 -|Typ 1 Umkehroperationen|Typ 2 Ausklammern|Typ 3 Substitution
95 -|{{formula}}x^2 = 2{{/formula}}|{{formula}}x^2-2x = 0{{/formula}}|{{formula}}x^4-40x^2+144 = 0{{/formula}}
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}}|{{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}}
89 +{{aufgabe id="Gleichungstypen einstudieren" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="5"}}
90 +Nenne eine passende Gleichung. Die Gleichung kann ich nach x auflösen, indem ich {{formula}} \ldots {{/formula}}
91 +(% class="abc" %)
92 +1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
93 +1. {{formula}} \ldots {{/formula}} von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
94 +1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
99 99  {{/aufgabe}}
100 100  
97 +
101 101  {{aufgabe id="Exponentialgleichungen (Logarithmieren)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="15"}}
102 102  Bestimme die Lösungsmenge der Exponentialgleichung:
103 103  (% class="abc" %)
2^-xund8.ggb
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