Lösung Gleichungstypen einstudieren

Typ 1: Umkehroperationen

1. \(x=\pm \sqrt{2}\)
2. \(x= \pm \sqrt[4]{e}\)
3. \(x = 1\) durch Exponentenvergleich
4. \(3e^x = \frac{1}{2}e^{-x}\qquad \mid \cdot e^x\)
   \(3e^{2x} = \frac{1}{2} \qquad \mid :3\)
   \(2x= ln\left( \frac{1}{6} \right)\qquad \mid :2 \)
   \(x= \frac{1}{2}ln\left( \frac{1}{6} \right) \)

Typ 2: Ausklammern

1. \(x^2-2x = 0\)
   \(x\cdot (x-2) = 0\)
   \(x_1=0 \quad , \quad x_2=2\)
2. \(x^e \cdot(2-x^e)=0\)
   \(x_1=0 \quad , \quad x_2=\sqrt[e]{2}\)
3. \(e^x \cdot (2-e^x) = 0\)
   \(e^x \neq 0 \quad , \quad x_1=ln(2)\)
4. \(3^x\cdot (x+4) = 0\)
   \(3^x \neq 0 \quad , \quad x_1=-4)\)

Typ 2: Substitution

1. \(x^4-40x^2+144 = 0\)
   Substitution \(x^2 = u\)
   \(u_1=-4 \quad , \quad u_2=36\)
   Resubstitution:
   \(x^2 = 4 \longleftrightarrow x_{1,2}=\pm2 \)
   \(x^2 = 36 \longleftrightarrow x_{3,4}=\pm6\)
     
2. \(x^{2e}+x^e+1 = 0\)
   Substitution \(x^e = u\)
   abc-Formel liefert keine Lösung, da die Diskriminante negativ ist.
     
3. \(10^{6x}-2\cdot 10^{3x}+1 = 0\) durch Exponentenvergleich
   Substitution \(10^{3x} = u\)
   \(u=1\)
   Resubstitution:
   \(10^{3x}=1 \longleftrightarrow x_{1}=0\)
     
4. \(3e^x-1 = \frac{1}{3}e^{-x}\)
   \(3e^x-1-\frac{1}{3}e^{-x}=0\)
   \(e^{-x}\cdot(3e^{2x}-e^x-\frac{1}{3})\)
   Substitution \(e^x = u\)
   \(3u^2-u - \frac{1}{3}=0\)
   \(u_1=\frac{1+\sqrt{5}}{6} \quad , \quad u_2=\frac{1-\sqrt{5}}{6}\)
   Resubstitution:
   \(e^x = \frac{1+\sqrt{5}}{6} \longleftrightarrow x_1=ln \left( \frac{1+\sqrt{5}}{6} \right)\)
   \(e^x = \frac{1-\sqrt{5}}{6} \longleftrightarrow \) keine weitere Lösung