1.8 Lösning 5a

\(\begin{align} \ln\,x & = 1 + \ln\,(x-1) \; & &\;| \; - \ln\,(x-1) \\ \ln\,x - \ln\,(x-1) & = 1 \; & &: \;\text{Logaritmlag 2 i VL} \\ \ln\,\left({x \over x-1}\right) & = 1 \; & &\;| \; e\,^{\cdot} \\ {x \over x-1} & = e \; & &\;| \; \cdot (x-1) \\ x & = e \cdot (x-1) \\ x & = e \cdot x - e \; & &\;| \; + e - x \\ e & = e \cdot x - x \; & &: \;\text{Bryt ut} \; x \;\text{i HL } \\ e & = x \cdot (e - 1) \; & &\;| \; / \; (e-1) \\ x & = {e \over e-1} \end{align}\)