Skillnad mellan versioner av "1.8 Lösning 5a"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 3: | Rad 3: | ||
\ln\,\left({x \over x-1}\right) & = 1 \; & &\;| \; e\,^{\cdot} \\ | \ln\,\left({x \over x-1}\right) & = 1 \; & &\;| \; e\,^{\cdot} \\ | ||
{x \over x-1} & = e \; & &\;| \; \cdot (x-1) \\ | {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}</math> | \end{align}</math> | ||
Nuvarande version från 20 april 2011 kl. 16.50
\(\begin{align} \ln\,x & = 1 + \ln\,(x-1) \; & &\;| \; - \ln\,(x-1) \\ \ln\,x - \ln\,(x-1) & = 1 \; & &: \;\text{Logaritmlag 2 i VL} \\ \ln\,\left({x \over x-1}\right) & = 1 \; & &\;| \; e\,^{\cdot} \\ {x \over x-1} & = e \; & &\;| \; \cdot (x-1) \\ x & = e \cdot (x-1) \\ x & = e \cdot x - e \; & &\;| \; + e - x \\ e & = e \cdot x - x \; & &: \;\text{Bryt ut} \; x \;\text{i HL } \\ e & = x \cdot (e - 1) \; & &\;| \; / \; (e-1) \\ x & = {e \over e-1} \end{align}\)
