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