Skillnad mellan versioner av "1.8 Lösning 5c"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
||
| Rad 2: | Rad 2: | ||
\ln((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \quad & &: \;\text{Konjugatregeln i VL}\\ | \ln((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \quad & &: \;\text{Konjugatregeln i VL}\\ | ||
\ln(x^2-1) & = \ln\,\left({3 \over 4}\right) \quad & &| \;10\,^{\cdot}\\ | \ln(x^2-1) & = \ln\,\left({3 \over 4}\right) \quad & &| \;10\,^{\cdot}\\ | ||
| − | x^2-1) & = {3 \over 4} | + | 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}</math> | \end{align}</math> | ||
Versionen från 11 april 2011 kl. 05.43
\(\begin{align} \ln\,(x+1) + \ln\,(x-1) & = \ln 3 - \ln 4 \quad & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\ \ln((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \quad & &: \;\text{Konjugatregeln i VL}\\ \ln(x^2-1) & = \ln\,\left({3 \over 4}\right) \quad & &| \;10\,^{\cdot}\\ x^2-1) & = {3 \over 4} \\ x^2 & = {3 \over 4} + 1 \\ x^2 & = {7 \over 4} \\ x & = {1 \over 2} \, \sqrt{7} \end{align}\)
