Math Problem Solver

$$\frac{\ln\left( 218 \cdot -0.02 + 3.1 \right) \cdot 0.9}{0.9} = \log{\left(\frac{63}{50} \right)} + i \pi$$

$$\begin{aligned}\frac{\ln\left( 218 \cdot -0.02 + 3.1 \right) \cdot 0.9}{0.9}&=\log{\left(218 \cdot -1 \cdot \frac{2}{100} + 3 + \frac{1}{10} \right)} \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\log{\left(218 \cdot \frac{-1}{50} + 3 + \frac{1}{10} \right)} \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\log{\left(\frac{-218}{50} + 3 + \frac{1}{10} \right)} \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\log{\left(- \frac{109}{25} + 3 + \frac{1}{10} \right)} \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\log{\left(- \frac{34}{25} + \frac{1}{10} \right)} \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\log{\left(- \frac{63}{50} \right)} \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\left(\pi \cdot i + \log{\left(\frac{63}{50} \right)}\right) \cdot \frac{9}{10} \cdot \frac{1}{\frac{9}{10}}\\&=\left(\pi \cdot i + \log{\left(\frac{63}{50} \right)}\right) \cdot \frac{9}{9}\\&=\left(9 \cdot \log{\left(\frac{63}{50} \right)} + 9 \cdot \pi \cdot i\right) \cdot \frac{1}{9}\\&=\pi \cdot i + \log{\left(\frac{63}{50} \right)}\end{aligned}$$