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Änderungskommentar: Löschung des Anhangs 2^-xund8.ggb
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1 -XWiki.holgerengels
1 +XWiki.elkehallmanngmxde
Inhalt
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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 -{{aufgabe id="Exponentialgleichungen (Logarithmieren)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="15"}}
11 -Bestimme die Lösungsmenge der Exponentialgleichung:
12 -(% class="abc" %)
10 +Aufgaben:
11 +– Logarithmus: graphisches Ermitteln vs. Operator
12 +Lösen von Exponentialgleichungen:
13 +– Vokabelheft für Umkehroperationen
14 +– Umkehrung der Rechenoperationen (Logarithmieren!) zzgl. Grundrechenarten
15 +– Faktorisierung durch Ausklammern und Satz vom Nullprodukt zzgl. Grundrechenarten
16 +– Substitution (abc-Formel, pq-Formel, Typ I) zzgl. Grundrechenarten
17 +- Näherungslösungen
13 13  
14 -1. {{formula}} e^x=3 {{/formula}}
15 -1. {{formula}} 2e^x-4=8 {{/formula}}
16 -1. {{formula}} 2e^{-0.5x}=6{{/formula}}
17 -1. {{formula}} e^x=-5 {{/formula}}
18 -1. {{formula}} 4\cdot 5^x=100 {{/formula}}
19 -{{/aufgabe}}
19 +Gleichungen:
20 +x+y = e --> y = e - x
21 +x*y = e --> y = e / x
22 +e^y = x --> y = ln(x)
20 20  
21 -{{aufgabe id="Exponentialgleichungen (Satz vom Nullprodukt)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="12"}}
22 -Bestimme die Lösungsmenge der Gleichung:
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}}
23 23  (% class="abc" %)
24 -1. {{formula}} 2x-x^{2}=0 {{/formula}}
25 -1. {{formula}} 2e^x-e^{2x}=0 {{/formula}}
26 -1. {{formula}} \frac{1}{3}e^x=e^{2x} {{/formula}}
27 -1. {{formula}} 3e^{-x}=2e^{2x} {{/formula}}
28 -1. {{formula}} 2x^e=x^{2e} {{/formula}}
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.
29 29  {{/aufgabe}}
30 30  
31 -{{aufgabe id="Exponentialgleichungen (Substitution)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="12"}}
32 -Bestimme die Lösungsmenge der Gleichung:
33 -(% class="abc" %)
34 -1. {{formula}} x^{2}-2x-3=0 {{/formula}}
35 -1. {{formula}} e^{2x}-2e^x-3=0 {{/formula}}
36 -1. {{formula}} e^x-2e^{\frac{1}{2}x}-3=0 {{/formula}}
37 -1. {{formula}} e^x-2-\frac{8}{e^x}}=0 {{/formula}}
38 -1. {{formula}} 2e^{4x}=e^{2x}+3 {{/formula}}
34 +{{aufgabe id="Gleichungen aufstellen II" afb="I" kompetenzen="K2,K5" quelle="Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
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}}
39 39  {{/aufgabe}}
40 40  
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:2^-xund8.svg||width="200px"]]
51 +|{{formula}} 2^{-x} = 8 {{/formula}}|{{formula}} x = \log_{2}(8) {{/formula}} |(((
52 +|x|0|1|2|3
53 +|y|0|1|8|27
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 +
61 +{{/aufgabe}}
62 +
41 41  {{aufgabe id="Logarithmen auswerten" afb="II" kompetenzen="K4,K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
42 42  Ordne (ohne WTR!) die Terme ihren Werten gemäß den Kästchen über dem Zahlenstrahl zu. Trage dafür die jeweiligen Buchstaben in die Kästchen ein.
43 43  
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55 55  1. {{formula}} \log_{10}(10) {{/formula}}
56 56  {{/aufgabe}}
57 57  
58 -{{aufgabe id="Exponentialgleichungen lösen (graphisch versus rechnerisch)" afb="I" kompetenzen="K4,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"}}
59 59  (% class="abc" %)
60 -Ermittle die Lösung der Gleichung {{formula}} e^x = 5 {{/formula}} graphisch und rechnerisch.
82 +Ermittle die Lösung der Gleichung {{formula}} 2^x = 5 {{/formula}} graphisch und rechnerisch.
61 61  {{/aufgabe}}
62 62  
63 -{{aufgabe id="Gleichungen aufstellen I" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="5"}}
64 -Nenne jeweils eine passende Gleichung:
65 -
66 -Die Gleichung kann ich nach x auflösen, indem ich …
85 +{{aufgabe id="Exponentialgleichungen Lösbarkeit (graphisch versus rechnerisch)" afb="I" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
67 67  (% class="abc" %)
68 -1. … die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
69 -1. … von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
70 -1. … die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
87 +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}}.
88 +{{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:. {{/formula}}
71 71  {{/aufgabe}}
72 72  
73 -{{aufgabe id="Darstellungen zuordnen" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="6"}}
74 -Ordne zu:
75 -(% class="border slim" %)
76 -|Implizite Gleichungen|Explizite Gleichungen|Wertetabellen|Schaubilder
77 -|{{formula}} x^{-3} = 8 {{/formula}}|{{formula}} x = \frac{1}{\sqrt[3]{8}} {{/formula}}|(((
78 -|x|0|1|2|3
79 -|y|1|2|4|8
80 -)))|[[image:2^xund8.svg||width="200px"]]
81 -|{{formula}} 2^x = 8 {{/formula}}|{{formula}} x = -\log_{2}(8) {{/formula}} |(((
82 -|x|0|1|2|3
83 -|y|n.d.|1|{{formula}}\frac{1}{8}{{/formula}}|{{formula}}\frac{1}{27}{{/formula}}
84 -)))|[[image:2^-xund8.svg||width="200px"]]
85 -|{{formula}} 2^{-x} = 8 {{/formula}}|{{formula}} x = \log_{2}(8) {{/formula}} |(((
86 -|x|0|1|2|3
87 -|y|1|{{formula}}\frac{1}{2}{{/formula}}|{{formula}}\frac{1}{4}{{/formula}}|{{formula}}\frac{1}{8}{{/formula}}
88 -)))|[[image:x^-3und8.svg||width="200px"]]
89 -{{/aufgabe}}
90 90  
91 -{{aufgabe id="Gleichungen gemeinsamer Form" afb="III" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="6"}}
92 -Die Gleichungen sehen auf den ersten Blick unterschiedlich aus, weisen aber ähnliche Strukturen auf und können alle mithilfe der Substitution gelöst werden. Selbstverständlich gibt es für manche Teilaufgaben auch andere Lösungswege ohne Substitution.
93 -(%class="abc"%)
94 -1. (((
95 -(%class="border slim"%)
96 -|(%align="center" width="160"%){{formula}}e^{2x}-4e^x+3=0{{/formula}}
97 -
98 -{{formula}}u:=\_\_\_{{/formula}}
99 -⬊|(%align="center" width="160"%){{formula}}x^{2e}-4x^e+3=0{{/formula}}
100 -
101 -{{formula}}u:=\_\_\_{{/formula}}
102 -🠗|(%align="center" width="160"%){{formula}}x^{-2}-4x^{-1}+3=0{{/formula}}
103 -
104 -{{formula}}u:=\_\_\_{{/formula}}
105 -⬋
106 -||(%align="center"%){{formula}}u^2-4u+3=0{{/formula}}
107 -(((
108 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
109 -|
110 -
111 -
112 -)))
113 -
114 -{{formula}}u_1=\_\_\_\quad;\quad u_2=\_\_\_{{/formula}}|
115 -|(%align="center"%)(((⬋
116 -{{formula}}\_\_\_:=u{{/formula}}
117 -(((
118 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
119 -|
120 -
121 -
122 -)))
123 -)))|(%align="center"%)(((🠗
124 -{{formula}}\_\_\_:=u{{/formula}}
125 -(((
126 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
127 -|
128 -
129 -
130 -)))
131 -)))|(%align="center"%)(((⬊
132 -{{formula}}\_\_\_:=u{{/formula}}
133 -(((
134 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
135 -|
136 -
137 -
138 -)))
139 -)))
140 -)))
141 -1. (((
142 -(%class="border slim"%)
143 -|(%align="center" width="160"%){{formula}}x^{-2}-3x^{-1}=0{{/formula}}
144 -
145 -{{formula}}u:=\_\_\_{{/formula}}
146 -⬊|(%align="center" width="160"%){{formula}}x^{2e}-3x^e=0{{/formula}}
147 -
148 -{{formula}}u:=\_\_\_{{/formula}}
149 -🠗|(%align="center" width="160"%){{formula}}e^{2x}-3e^x=0{{/formula}}
150 -
151 -{{formula}}u:=\_\_\_{{/formula}}
152 -⬋
153 -||(%align="center"%){{formula}}u^2-3u=0{{/formula}}
154 -(((
155 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
156 -|
157 -
158 -
159 -)))
160 -
161 -{{formula}}u_1=\_\_\_\quad;\quad u_2=\_\_\_{{/formula}}|
162 -|(%align="center"%)(((⬋
163 -{{formula}}\_\_\_:=u{{/formula}}
164 -(((
165 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
166 -|
167 -
168 -
169 -)))
170 -)))|(%align="center"%)(((🠗
171 -{{formula}}\_\_\_:=u{{/formula}}
172 -(((
173 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
174 -|
175 -
176 -
177 -)))
178 -)))|(%align="center"%)(((⬊
179 -{{formula}}\_\_\_:=u{{/formula}}
180 -(((
181 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
182 -|
183 -
184 -
185 -)))
186 -)))
187 -)))
188 -1. (((
189 -(%class="border slim"%)
190 -|(%align="center" width="160"%){{formula}}x^{-2}-2x^{-1}+3=0{{/formula}}
191 -
192 -{{formula}}u:=\_\_\_{{/formula}}
193 -⬊|(%align="center" width="160"%){{formula}}x^{2e}-2x^e+3=0{{/formula}}
194 -
195 -{{formula}}u:=\_\_\_{{/formula}}
196 -🠗|(%align="center" width="160"%){{formula}}e^{2x}-2e^x+3=0{{/formula}}
197 -
198 -{{formula}}u:=\_\_\_{{/formula}}
199 -⬋
200 -||(%align="center"%){{formula}}u^2-2u+3=0{{/formula}}
201 -(((
202 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
203 -|
204 -
205 -
206 -)))
207 -
208 -{{formula}}u_1=\_\_\_\quad;\quad u_2=\_\_\_{{/formula}}|
209 -|(%align="center"%)(((⬋
210 -{{formula}}\_\_\_:=u{{/formula}}
211 -(((
212 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
213 -|
214 -
215 -
216 -)))
217 -)))|(%align="center"%)(((🠗
218 -{{formula}}\_\_\_:=u{{/formula}}
219 -(((
220 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
221 -|
222 -
223 -
224 -)))
225 -)))|(%align="center"%)(((⬊
226 -{{formula}}\_\_\_:=u{{/formula}}
227 -(((
228 -(%class="border slim" style="width: 100%; margin-bottom: 0px"%)
229 -|
230 -
231 -
232 -)))
233 -)))
234 -)))
235 -{{/aufgabe}}
236 -
237 -{{aufgabe id="Gleichungstypen einstudieren" afb="III" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="20"}}
92 +{{aufgabe id="Gleichungstypen einstudieren" afb="II" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe, Martina Wagner" cc="BY-SA" zeit="20"}}
238 238  Bestimme die Lösung der folgenden Gleichungen:
239 239  
240 240  (% class="border slim " %)
241 -|Typ 1 (Umkehroperationen)|Typ 2 (Ausklammern)|Typ 3 (Substitution)
96 +|Typ 1 Umkehroperationen|Typ 2 Ausklammern|Typ 3 Substitution
242 242  |{{formula}}x^2 = 2{{/formula}}|{{formula}}x^2-2x = 0{{/formula}}|{{formula}}x^4-40x^2+144 = 0{{/formula}}
243 -|{{formula}}x^4 = e{{/formula}}|{{formula}}2x^e = x^{2e}{{/formula}}|{{formula}}x^{2e}+x^e+1 = 0{{/formula}}
98 +|{{formula}}x^4 = e{{/formula}}|{{formula}}2x^e = x^{2e}{{/formula}}|{{formula}}x^{2x}+x^e+1 = 0{{/formula}}
244 244  |{{formula}}e^x = e{{/formula}}|{{formula}}2e^x = e^{2x}{{/formula}}|{{formula}}10^{6x}-2\cdot 10^{3x}+1 = 0{{/formula}}
245 245  |{{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}}
246 246  {{/aufgabe}}
247 247  
248 -{{aufgabe id=" Exponentialgleichungen rückwärts lösen" afb="II" kompetenzen="K2,K5" quelle="Martina Wagner" lizenz="BY-SA"}}
103 +Nenne eine passende Gleichung. Die Gleichung kann ich nach x auflösen, indem ich {{formula}} \ldots {{/formula}}
249 249  (% class="abc" %)
250 -1. ((({{{ }}}
105 +1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten durch 5 dividiere und damit die Lösung {{formula}} x = \frac{2}{5} {{/formula}} erhalte.
106 +1. {{formula}} \ldots {{/formula}} von beiden Termen die 5-te Wurzel ziehe und damit die Lösung {{formula}} x = \sqrt[5]{2} {{/formula}} erhalte.
107 +1. {{formula}} \ldots {{/formula}} die Terme auf beiden Seiten zur Basis 5 logarithmiere und damit die Lösung {{formula}} x = \log_5(2) {{/formula}} erhalte.
108 +{{/aufgabe}}
251 251  
252 -{{formula}}
253 -\begin{align*}
254 -\square e^x-2 &= 0\\
255 -\square e^x &=\square\quad \left|:\square\\
256 -e^x &= \square \\
257 -x &= 0
258 -\end{align*}
259 -{{/formula}}
260 -)))
261 -1. ((({{{ }}}
262 262  
263 -{{formula}}
264 -\begin{align*}
265 -e^{2x}-\square e^x &= 0 \\
266 -e^x \cdot (\square-\square) &= 0 \left|\left| \text{ SVNP }
267 -\end{align*}
268 -{{/formula}}
111 +{{aufgabe id="Exponentialgleichungen (Logarithmieren)" afb="I" kompetenzen="K5" quelle="Elke Hallmann, Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="15"}}
112 +Bestimme die Lösungsmenge der Exponentialgleichung:
113 +(% class="abc" %)
114 +1. {{formula}} 4\cdot 0,5^x=100 {{/formula}}
115 +1. {{formula}} e^x=3 {{/formula}}
116 +1. {{formula}} 2e^x-4=8 {{/formula}}
117 +1. {{formula}} 2e^{-0.5x}=6{{/formula}}
118 +1. {{formula}} e^x=-5 {{/formula}}
119 +{{/aufgabe}}
269 269  
270 -{{formula}}
271 -e^x \neq 0 ~und~ e^x-\square = 0{{/formula}}
272 -{{formula}} e^x=\square {{/formula}}
273 -{{formula}} x =\square {{/formula}}
274 -)))
275 -1. ((({{{ }}}
121 +{{aufgabe id="Exponentialgleichungen (Nullproduktsatz)" afb="II" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="12"}}
122 +Bestimme die Lösungsmenge der Gleichung:
123 +(% class="abc" %)
124 +1. {{formula}} 2x=x^{2} {{/formula}}
125 +1. {{formula}} 2x^e=x^{2e} {{/formula}}
126 +1. {{formula}} 2e^x=e^{2x} {{/formula}}
127 +{{/aufgabe}}
276 276  
277 -{{formula}}
278 -\begin{align*}
279 -e^{2x}-\square e^x+\square &= 0 \quad \left|\left|\text{ Subst.: } e^x:=\square\\
280 -z^2-\square z + \square &= 0 \quad \left|\left|\text{ Mitternachtsformel/abc-Formel } &
281 -\end{align*}
282 -{{/formula}}
129 +{{aufgabe id="Exponentialgleichungen (Substitution)" afb="III" kompetenzen="K5" quelle="Martin Rathgeb" cc="BY-SA" zeit="12"}}
130 +Bestimme die Lösungsmenge der Gleichung:
131 +(% class="abc" %)
132 +1. {{formula}} 2x-3=x^{2} {{/formula}}
133 +1. {{formula}} 2x^e-3=x^{2e} {{/formula}}
134 +1. {{formula}} 2e^x-3=e^{2x} {{/formula}}
135 +1. {{formula}} 2e^{x-3}=e^{2x-3} {{/formula}}
136 +{{/aufgabe}}
283 283  
284 -{{formula}}
285 -\begin{align*}
286 -\Rightarrow z_{1,2}&=\frac{\square\pm\sqrt{\square^2-4\cdot\square\cdot\square}}{2\cdot\square}\\
287 -z_{1,2}&=\frac{\square+\square}{\square}
288 -\end{align*}
289 -{{/formula}}
290 -
291 -{{formula}}
292 -\begin{align*}
293 -&\text{Resubst.: } z:= e^x\\
294 -&e^x=\square \Rightarrow x \approx 0,693147...\\
295 -\end{align*}
296 -{{/formula}}
297 -)))
138 +{{aufgabe id="Exponentialgleichungen" afb="I" kompetenzen="K5" quelle="Niklas Wunder" cc="BY-SA" zeit="5"}}
139 +Bestimme die Lösungsmenge der folgenden Exponentialgleichungen
140 +(% class="abc" %)
141 +1. {{formula}} 3^{x+1}=81 {{/formula}}
142 +1. {{formula}} 5^{2x}=25^{2x+2} {{/formula}}
143 +1. {{formula}} 10^{x}=500{{/formula}}
144 +1. {{formula}} 2^{x+3}=4^{x-1} {{/formula}}
298 298  {{/aufgabe}}
299 299  
300 -{{aufgabe id="Gleichungen aufstellen II" afb="III" kompetenzen="K2,K5" quelle="Martin Rathgeb, Dirk Tebbe" cc="BY-SA" zeit="10"}}
301 -Nenne möglichst viele (wahre) Gleichungen der folgenden Formen, wobei {{formula}} a, b, c \in \{2; 3; 4; \ldots; 16\} {{/formula}} gelten soll:
302 -{{formula}} c = a^b\:; \qquad c = \sqrt[a]{b}\:; \qquad c = \log_a(b)\:; \qquad c = a\cdot b\:. {{/formula}}
147 +{{aufgabe id="Exponentialgleichungen graphisch" afb="II" kompetenzen="K4,K6" quelle="Niklas Wunder" cc="BY-SA" zeit="5"}}
148 +Löse mit Hilfe der nebenstehenden Abbildung folgende Exponentialgleichungen näherungsweise. Hinweis: Ordne die linke und die rechte Seite der jeweiligen Gleichung passend den Funktionsgraphen zu.
149 +(% class="abc" %)
150 +1. {{formula}} 2^x=(\frac{3}{4})^x+2 {{/formula}}
151 +1. {{formula}} 7-e^{x-3}=(\frac{3}{4})^x+2 {{/formula}}
152 +1. {{formula}} 2^x=1{,}5^{x+2}-0{,}5 {{/formula}}
153 +1. {{formula}} 7-e^{x-3}=4-\frac{1}{2}\,x {{/formula}}
154 +
155 +[[image:ExpGlei.svg||width="600px"]]
303 303  {{/aufgabe}}
304 304  
305 305  {{seitenreflexion/}}
2^-xund8.svg
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1 -XWiki.holgerengels
Größe
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BPE 4.5 A Gleichungen Gemeinsamer Form.pdf
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Darstellungen zuordnen.ggb
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