Skillnad mellan versioner av "1.8 Lösning 5a"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
<math>\begin{align} \ln\,x & = 1 + \ln\,(x-1) \; & &\;| \; - \ln\,(x-1) \\ | <math>\begin{align} \ln\,x & = 1 + \ln\,(x-1) \; & &\;| \; - \ln\,(x-1) \\ | ||
\ln\,x - \ln\,(x-1) & = 1 \; & &: \;\text{Logaritmlag 2 i VL} \\ | \ln\,x - \ln\,(x-1) & = 1 \; & &: \;\text{Logaritmlag 2 i VL} \\ | ||
| − | \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-1) \\ |
| − | x & = e \cdot x - e \; & &\;| \; + e - x | + | x & = e \cdot x - e \; & &\;| \; + e \\ |
| + | x + e & = e \cdot x \; & &\;| \; + e - x \\ | ||
e & = e \cdot x - x \; & &: \;\text{Bryt ut} \; x \;\text{i HL } \\ | e & = e \cdot x - x \; & &: \;\text{Bryt ut} \; x \;\text{i HL } \\ | ||
e & = x \cdot (e - 1) \; & &\;| \; / \; (e-1) \\ | e & = x \cdot (e - 1) \; & &\;| \; / \; (e-1) \\ | ||
x & = {e \over e-1} | x & = {e \over e-1} | ||
\end{align}</math> | \end{align}</math> | ||
Versionen från 9 oktober 2015 kl. 10.54
\(\begin{align} \ln\,x & = 1 + \ln\,(x-1) \; & &\;| \; - \ln\,(x-1) \\ \ln\,x - \ln\,(x-1) & = 1 \; & &: \;\text{Logaritmlag 2 i VL} \\ \ln\,\left({x \over x-1}\right) & = 1 \; & &\;| \; e\,^{\cdot} \\ {x \over x-1} & = e \; & &\;| \; \cdot (x-1) \\ x & = e \cdot (x-1) \\ x & = e \cdot x - e \; & &\;| \; + e \\ x + e & = e \cdot x \; & &\;| \; + e - x \\ e & = e \cdot x - x \; & &: \;\text{Bryt ut} \; x \;\text{i HL } \\ e & = x \cdot (e - 1) \; & &\;| \; / \; (e-1) \\ x & = {e \over e-1} \end{align}\)
