Skillnad mellan versioner av "1.8 Lösning 5c"
Från Mathonline
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 3 - \ln 4 \; & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\ | <math>\begin{align} \ln\,(x+1) + \ln\,(x-1) & = \ln 3 - \ln 4 \; & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\ | ||
\ln\,((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\ | \ln\,((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\ | ||
| − | \ln\,(x^2-1) & = \ln\,\left({3 \over 4}\right) \; & &| \; | + | \ln\,(x^2-1) & = \ln\,\left({3 \over 4}\right) \; & &| \;e\,^{\cdot}\\ |
(x^2-1) & = {3 \over 4} \\ | (x^2-1) & = {3 \over 4} \\ | ||
x^2 & = {3 \over 4} + 1 \\ | x^2 & = {3 \over 4} + 1 \\ | ||
Versionen från 11 april 2011 kl. 09.30
\(\begin{align} \ln\,(x+1) + \ln\,(x-1) & = \ln 3 - \ln 4 \; & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\ \ln\,((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\ \ln\,(x^2-1) & = \ln\,\left({3 \over 4}\right) \; & &| \;e\,^{\cdot}\\ (x^2-1) & = {3 \over 4} \\ x^2 & = {3 \over 4} + 1 \\ x^2 & = {7 \over 4} \\ x & = {1 \over 2} \, \sqrt{7} \end{align}\)
