Änderungen von Dokument Lösung Exponentialgleichungen (Substitution)

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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!