Änderungen von Dokument Lösung Exponentialgleichungen (Logarithmieren)
Zuletzt geändert von akukin am 2025/08/11 15:31
Zusammenfassung
-
Seiteneigenschaften (2 geändert, 0 hinzugefügt, 0 gelöscht)
Details
- Seiteneigenschaften
-
- Dokument-Autor
-
... ... @@ -1,1 +1,1 @@ 1 -XWiki. fujan1 +XWiki.akukin - Inhalt
-
... ... @@ -1,23 +1,28 @@ 1 -(% class=" abc" %)2 -1. {{formula}} e^x=3 \quad \ left|ln( ) {{/formula}}1 +(% class="123" %) 2 +1. {{formula}} e^x=3 \quad \mid ln( ) {{/formula}} 3 3 {{formula}} ln(e^x)=ln(3) {{/formula}} 4 4 {{formula}} x=ln(3) {{/formula}} 5 -{{formula}}\mathbb{L} {{/formula}} 6 - 7 - 8 -1. {{formula}} 2e^x-4=8 \quad \left|+4{{/formula}} 9 -{{formula}} 2e^x=12 \quad \left|:2{{/formula}} 10 -{{formula}} e^x=6 \quad \left| ln( ) {{/formula}} 5 +{{formula}}\mathbb{L}= \left\{ ln(3) \right\} {{/formula}} 6 + 7 +1. {{formula}} 2e^x-4=8 \quad \mid +4{{/formula}} 8 +{{formula}} 2e^x=12 \quad \mid :2{{/formula}} 9 +{{formula}} e^x=6 \quad \mid ln( ) {{/formula}} 11 11 {{formula}} x=ln(6) {{/formula}} 12 - 13 -1. {{formula}} 2e^{-0.5x}=6 \quad \left|:2 {{/formula}} 14 -{{formula}} e^{-0.5x}=3 \quad \left| ln( ) {{/formula}} 15 -{{formula}} -0.5x=ln(3) \quad \left|\cdot (-2) {{/formula}} 11 +{{formula}}\mathbb{L}= \left\{ ln(6) \right\} {{/formula}} 12 + 13 +1. {{formula}} 2e^{-0.5x}=6 \quad \mid :2 {{/formula}} 14 +{{formula}} e^{-0.5x}=3 \quad \mid ln( ) {{/formula}} 15 +{{formula}} -0.5x=ln(3) \quad \mid \cdot (-2) {{/formula}} 16 16 {{formula}} x=-2 \cdot ln(3) {{/formula}} 17 - 18 -1. {{formula}} e^x=-5 \quad \left| ln( ) {{/formula}} 17 +{{formula}}\mathbb{L}= \left\{-2 \cdot ln(3) \right\} {{/formula}} 18 + 19 +1. {{formula}} e^x=-5 \quad \mid ln( ) {{/formula}} 19 19 {{formula}}x=ln(-5){{/formula}} 20 20 keine Lösung! 21 - 22 -1. {{formula}} 4\cdot 5^x=100 {{/formula}} 23 - 22 +{{formula}}\mathbb{L}= \left\{ \right\} {{/formula}} 23 + 24 +1. {{formula}} 4\cdot 5^x=100 \quad \mid:4 {{/formula}} 25 +{{formula}} 5^x=25 \quad \mid \text{Exponentenvergleich} {{/formula}} 26 +{{formula}} 5^x=5^2 {{/formula}} 27 +{{formula}} x=2 {{/formula}} 28 +{{formula}}\mathbb{L}= \left\{ 2 \right\} {{/formula}}