Wiki-Quellcode von Lösung Exponentialgleichungen lösen
Zuletzt geändert von Christoph Gommel am 2026/02/02 22:17
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
4.1 | 1 | |
| 2 | a) | ||
| 3 | {{formula}} | ||
| |
3.1 | 4 | \begin{align*} |
| |
4.1 | 5 | 4\cdot 5^x+20=120 &&& \mid -20 \\ |
| 6 | 4\cdot 5^x=100 &&& \mid :4 \\ | ||
| |
5.1 | 7 | 5^x=25 &&& \\ |
| |
7.1 | 8 | 5^x=5^2 &&& \mid Koeffizientenvergleich \\ |
| |
5.1 | 9 | x=2 \\ |
| 10 | \mathbb{L}= \left\{ 2 \right\} | ||
| |
4.1 | 11 | \end{align*} |
| |
7.1 | 12 | {{/formula}} |
| |
5.1 | 13 | |
| 14 | b) | ||
| 15 | {{formula}} | ||
| 16 | \begin{align*} | ||
| 17 | 2\cdot (2^x+4)=8 &&& \mid :2 \\ | ||
| |
7.1 | 18 | 2^x+4=4 &&& \mid -4 \\ |
| 19 | 2^x=0 &&& \\ | ||
| 20 | \mathbb{L}= \left\{ \right\} | ||
| |
5.1 | 21 | \end{align*} |
| |
7.1 | 22 | {{/formula}} |
| 23 | |||
| 24 | c) | ||
| 25 | {{formula}} | ||
| 26 | \begin{align*} | ||
| |
24.1 | 27 | -2\cdot 3^x=-6 &&& \mid :(-2) \\ |
| |
26.1 | 28 | 3^x=3^1 &&& \mid Koeffizientenvergleich \\ |
| |
7.1 | 29 | \mathbb{L}= \left\{ 1 \right\} |
| 30 | \end{align*} | ||
| 31 | {{/formula}} | ||
| |
8.1 | 32 | |
| 33 | d) | ||
| 34 | {{formula}} | ||
| 35 | \begin{align*} | ||
| |
31.1 | 36 | 1+2^x=7 &&&& \mid -1 \\ |
| |
30.1 | 37 | \: 2^x=6 &&&& \mid log_2 \\ |
| |
41.1 | 38 | \; x=log_2 6 \\ |
| |
34.1 | 39 | \quad \mathbb{L}= \left\{ 2,58 \right\} |
| |
8.1 | 40 | \end{align*} |
| 41 | {{/formula}} | ||
| |
22.1 | 42 | |
| 43 | e) | ||
| 44 | {{formula}} | ||
| 45 | \begin{align*} | ||
| 46 | 3\cdot (5-3^x)=-21 &&& \mid :3 \\ | ||
| 47 | 5-3^x=-7 &&& \mid -5 \\ | ||
| 48 | -3^x=-12 &&& \mid \cdot (-1)\\ | ||
| 49 | 3^x=12 &&& \mid log_3 \\ | ||
| 50 | x=log_3 12 \\ | ||
| |
23.1 | 51 | \mathbb{L}= \left\{ 2,26 \right\} |
| |
22.1 | 52 | \end{align*} |
| 53 | {{/formula}} |