Wiki-Quellcode von Lösung Negative Exponenten Erklärung
Version 1.1 von Tina Müller am 2024/10/14 14:32
Zeige letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
| 1 | Zu zeigen ist: {{formula}}\(2^{-2} = \frac{1}{4}\){{/formula}} mithilfe des Potenzgesetzes | ||
| 2 | {{formula}} | ||
| 3 | \[ | ||
| 4 | \frac{a^n}{a^m} = a^{n-m} | ||
| 5 | \] | ||
| 6 | {{/formula}} | ||
| 7 | Wir setzen {{formula}}\(n - m = -2\){{/formula}} . Eine einfache Wahl ist: | ||
| 8 | {{formula}} | ||
| 9 | \(n = 0\) | ||
| 10 | \(m = 2\){{/formula}} | ||
| 11 | |||
| 12 | Dann gilt: | ||
| 13 | {{formula}} | ||
| 14 | \[ | ||
| 15 | n - m = 0 - 2 = -2 | ||
| 16 | \]{{/formula}} | ||
| 17 | |||
| 18 | Jetzt wenden wir das Potenzgesetz an: | ||
| 19 | {{formula}} | ||
| 20 | \[ | ||
| 21 | \frac{a^0}{a^2} = a^{0-2} = a^{-2} | ||
| 22 | \]{{/formula}} | ||
| 23 | |||
| 24 | Setzen wir {{formula}}\(a = 2\) {{/formula}}ein: | ||
| 25 | {{formula}} | ||
| 26 | \[ | ||
| 27 | \frac{2^0}{2^2} = 2^{-2} | ||
| 28 | \]{{/formula}} | ||
| 29 | |||
| 30 | Da {{formula}}\(2^0 = 1\){{/formula}} und {{formula}}\(2^2 = 4\){{/formula}}, ergibt sich: | ||
| 31 | {{formula}} | ||
| 32 | \[ | ||
| 33 | \frac{1}{4} = 2^{-2} | ||
| 34 | \]{{/formula}} | ||
| 35 | |||
| 36 | und somit: | ||
| 37 | {{formula}} | ||
| 38 | \[ | ||
| 39 | 2^{-2} = \frac{1}{4} | ||
| 40 | \]{{/formula}} | ||
| 41 | |||
| 42 | Damit haben wir durch Anwendung der Potenzgesetze gezeigt, dass {{formula}}\(2^{-2} = \frac{1}{4}\){{/formula}} ist. |