Skillnad mellan versioner av "1.7 Lösning 5c"
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Taifun (Diskussion | bidrag) m (Created page with "<math>\begin{align} \lg\,(x+1) + \lg\,(x-1) & = \lg 3 - \lg 4 \; & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\ \lg\,((x+1) \cdot (x-1)) & = \lg\,...") |
Taifun (Diskussion | bidrag) m |
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| Rad 2: | Rad 2: | ||
\lg\,((x+1) \cdot (x-1)) & = \lg\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\ | \lg\,((x+1) \cdot (x-1)) & = \lg\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\ | ||
\lg\,(x^2-1) & = \lg\,\left({3 \over 4}\right) \; & &| \;10\,^{\cdot}\\ | \lg\,(x^2-1) & = \lg\,\left({3 \over 4}\right) \; & &| \;10\,^{\cdot}\\ | ||
| − | + | x^2 - 1 & = {3 \over 4} \\ | |
x^2 & = {3 \over 4} + 1 \\ | x^2 & = {3 \over 4} + 1 \\ | ||
x^2 & = {7 \over 4} \\ | x^2 & = {7 \over 4} \\ | ||
x & = {1 \over 2} \, \sqrt{7} | x & = {1 \over 2} \, \sqrt{7} | ||
\end{align}</math> | \end{align}</math> | ||
Versionen från 11 april 2011 kl. 13.41
\(\begin{align} \lg\,(x+1) + \lg\,(x-1) & = \lg 3 - \lg 4 \; & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\ \lg\,((x+1) \cdot (x-1)) & = \lg\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\ \lg\,(x^2-1) & = \lg\,\left({3 \over 4}\right) \; & &| \;10\,^{\cdot}\\ x^2 - 1 & = {3 \over 4} \\ x^2 & = {3 \over 4} + 1 \\ x^2 & = {7 \over 4} \\ x & = {1 \over 2} \, \sqrt{7} \end{align}\)
