\( { 7 }^{ x } + 20 = { 7 }^{ 2x+1 } \quad | -7^{x} -20 \\ 0 = 7^{2x} · 7^{1} - 7^{x} - 20 \)
\( z = 7^{x} \)
\( 0 = 7z^2 - z- 20 \quad |:7 \\ 0 = z^2 - \frac{1}{7} z - \frac{20}{7} \quad | \text{pq-Formel} \)
\( z_{1,2}=\frac{1}{14} \pm \sqrt{\frac{1}{196}+\frac{20}{7}} \)
\( z_{1,2}=\frac{1}{14} \pm \sqrt{\frac{1+560}{196}} \)
\( z_{1,2}=\frac{1}{14} \pm \sqrt{\frac{561}{196}} \)
\( z_{1,2}=\frac{1}{14} \pm \frac{\sqrt{561}}{14}=\frac{1}{14}(1 \pm \sqrt{561}) \)
\( z_{1} \approx 1,76 \)
\( z_{2} \approx-1,62 \)
Resubstitution:
\( z = 7^{x} \\ 1,76 = 7^{x} \quad | -1,62 = 7^{x} \rightarrow \text{ keine Lösung } \)
\( \ln(1,76) = x · ln(7) \\ x = \frac{\ln(1,76)}{ \ln(7) } \approx \frac{0,565}{1,95} \\ x \approx 0,29 \)
