Änderungen von Dokument Lösung Exponentialgleichungen (Logarithmieren)
Zuletzt geändert von Kim Fujan am 2025/05/20 11:11
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... ... @@ -1,14 +1,28 @@ 1 1 (% class="abc" %) 2 -1. {{formula}} e^x=3 \ square\left| ln( ) {{/formula}}2 +1. {{formula}} e^x=3 \quad \left| ln( ) {{/formula}} 3 3 {{formula}} ln(e^x)=ln(3) {{/formula}} 4 4 {{formula}} x=ln(3) {{/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}} 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}} 8 8 {{formula}} x=ln(6) {{/formula}} 9 - 1.{{formula}}2e^{-0.5x}=6{{/formula}}11 +{{formula}}\mathbb{L}= \left\{ ln(6) \right\} {{/formula}} 10 10 11 -1. {{formula}} e^x=-5 {{/formula}} 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}} 12 12 13 -1. {{formula}} 4\cdot 5^x=100 {{/formula}} 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}} 14 14 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}}