Änderungen von Dokument Lösung Exponentialgleichungen (Substitution)
Zuletzt geändert von Kim Fujan am 2025/05/21 10:47
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... ... @@ -1,12 +1,49 @@ 1 1 (% class="abc" %) 2 -1. {{formula}} x^{2}+2x-3=0 {{/formula}} 2 +1. {{formula}} x^{2}-2x-3=0 {{/formula}} 3 + 3 3 Lösung mit abc-Formel: 4 -{{formula}}x_{1,2}=\frac{2\pm \sqrt{4+12}}{2}{{/formula}} 5 - 6 -1. {{formula}} e^{2x}+2e^x-3=0 {{/formula}} 5 +{{formula}}x_{1,2}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}{{/formula}} 6 +{{formula}} x_{1}=3 \quad ; \quad x_{2}=-1 {{/formula}} 7 7 8 -1. {{formula}} e^x +2e^{\frac{1}{2}x}-3=0 {{/formula}}8 +1. {{formula}} e^{2x}-2e^x-3=0 {{/formula}} 9 9 10 +Substitution: {{formula}} e^x=u {{/formula}} 11 +{{formula}} u^{2}+2u-3=0 {{/formula}} 12 + 13 +Lösung mit abc-Formel: 14 +{{formula}}u_{1,2}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}{{/formula}} 15 +{{formula}} u_{1}=3 \quad ; \quad u_{2}=-1 {{/formula}} 16 + 17 +Resubstitution: 18 +{{formula}} e^x=3 \quad \Longleftrightarrow \quad x=ln(3) {{/formula}} 19 +{{formula}} e^x=-1 \quad \Longleftrightarrow \quad {{/formula}} keine weitere Lösung! 20 + 21 +1. {{formula}} e^x-2e^{\frac{1}{2}x}-3=0 {{/formula}} 22 + 23 +Substitution: {{formula}} e^{\frac{1}{2}x}=u {{/formula}} 24 +{{formula}} u^{2}+2u-3=0 {{/formula}} 25 + 26 +Lösung mit abc-Formel: 27 +{{formula}}u_{1,2}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}{{/formula}} 28 +{{formula}} u_{1}=3 \quad ; \quad u_{2}=-1 {{/formula}} 29 + 30 +Resubstitution: 31 +{{formula}} e^{\frac{1}{2}x}=3 \quad \Longleftrightarrow \quad x=2 \cdot ln(3) {{/formula}} 32 +{{formula}} e^{\frac{1}{2}x}=-1 \quad \Longleftrightarrow \quad {{/formula}} keine weitere Lösung! 33 + 10 10 1. {{formula}} e^x-2-\frac{15}{e^x}}=0 {{/formula}} 35 +{{formula}} e^{-x} \cdot (e^{2x}-2e^x-15)=0 {{/formula}} 11 11 12 12 1. {{formula}} 2e^{4x}=e^{2x}+3 {{/formula}} 38 +{{formula}} 2e^{4x}-e^{2x}-3=0 {{/formula}} 39 + 40 +Substitution: {{formula}} e^{2x}=u {{/formula}} 41 +{{formula}} 2u^{2}-u-3=0 {{/formula}} 42 + 43 +Lösung mit abc-Formel: 44 +{{formula}}u_{1,2}=\frac{1\pm \sqrt{1+24}}{2}=\frac{1\pm 5}{2}{{/formula}} 45 +{{formula}} u_{1}=3 \quad ; \quad u_{2}=-2 {{/formula}} 46 + 47 +Resubstitution: 48 +{{formula}} e^{\frac{1}{2}x}=3 \quad \Longleftrightarrow \quad x=2 \cdot ln(3) {{/formula}} 49 +{{formula}} e^{\frac{1}{2}x}=-2 \quad \Longleftrightarrow \quad {{/formula}} keine weitere Lösung!