Änderungen von Dokument Lösung Exponentialgleichungen (Logarithmieren)
Zuletzt geändert von Kim Fujan am 2025/05/20 11:11
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... ... @@ -1,28 +1,14 @@ 1 1 (% class="abc" %) 2 -1. {{formula}} e^x=3 \qua d\left| ln( ) {{/formula}}2 +1. {{formula}} e^x=3 \square\left| ln( ) {{/formula}} 3 3 {{formula}} ln(e^x)=ln(3) {{/formula}} 4 4 {{formula}} x=ln(3) {{/formula}} 5 -{{formula}}\mathbb{L}= \left\{ ln(3) \right\} {{/formula}} 6 - 7 -1. {{formula}} 2e^x-4=8 \quad \left|+4{{/formula}} 8 -{{formula}} 2e^x=12 \quad \left|:2{{/formula}} 9 -{{formula}} e^x=6 \quad \left| ln( ) {{/formula}} 5 +1. {{formula}} 2e^x-4=8 \square\ \left|+4{{/formula}} 6 +{{formula}} 2e^x=12 \square \left|:2{{/formula}} 7 +{{formula}} e^x=6 \square\ \left| ln( ) {{/formula}} 10 10 {{formula}} x=ln(6) {{/formula}} 11 -{{formula}} \mathbb{L}=\left\{ln(6) \right\}9 +1. {{formula}} 2e^{-0.5x}=6{{/formula}} 12 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}} 16 -{{formula}} x=-2 \cdot ln(3) {{/formula}} 17 -{{formula}}\mathbb{L}= \left\{-2 \cdot ln(3) \right\} {{/formula}} 11 +1. {{formula}} e^x=-5 {{/formula}} 18 18 19 -1. {{formula}} e^x=-5 \quad \left| ln( ) {{/formula}} 20 -{{formula}}x=ln(-5){{/formula}} 21 -keine Lösung! 22 -{{formula}}\mathbb{L}= \left\{ \right\} {{/formula}} 13 +1. {{formula}} 4\cdot 5^x=100 {{/formula}} 23 23 24 -1. {{formula}} 4\cdot 5^x=100 \quad \left|:4 {{/formula}} 25 -{{formula}} 5^x=25 \quad \left| \text{Exponentenvergleich} {{/formula}} 26 -{{formula}} 5^x=5^2 {{/formula}} 27 -{{formula}} x=2 {{/formula}} 28 -{{formula}}\mathbb{L}= \left\{ 2 \right\} {{/formula}}