TeX Quellcode:

\begin{array}{rclcl}\ln\left(4x^{\frac{7}{8}}\right)+\ln\left(3x^{-\frac{1}{8}}\right) &=& \dfrac{7}{4}\ln\left(\sqrt[7]{16}x\right) &\vert& \text{1. Logarithmengesetz} \\\\\ln\left(4x^{\frac{7}{8}}\cdot 3x^{-\frac{1}{8}}\right) &=& \dfrac{7}{4}\ln\left(\sqrt[7]{16}x\right) &\vert& \text{3. Logarithmengesetz} \\\ln\left(12x^{\frac{3}{4}}\right) &=& \ln\left(\left(\sqrt[7]{16}x\right)^{\frac{7}{4}}\right) \\\ln\left(12x^{\frac{3}{4}}\right) &=& \ln\left(\left(\sqrt[7]{16}\right)^{\frac{7}{4}}x^{\frac{7}{4}}\right) \\\ln\left(12x^{\frac{3}{4}}\right) &=& \ln\left(2x^{\frac{7}{4}}\right) &\vert& e^* \\12x^{\frac{3}{4}} &=& 2x^{\frac{7}{4}} &\vert& :2 \\6x^{\frac{3}{4}} &=& x^{\frac{7}{4}} &\vert& :x^{\frac{3}{4}} \\6 &=& x^{\frac{7}{4}-\frac{3}{4}} \\6 &=& x^{1} \\6 &=& x \in \mathbb{D} \\\\\mathbb{L} &=& \{6\}\end{array}