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1.8 Lösning 5c - Versionshistorik
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Taifun den 11 april 2011 kl. 13.46
2011-04-11T13:46:29Z
<p></p>
<p><b>Ny sida</b></p><div><math>\begin{align} \ln\,(x+1) + \ln\,(x-1) & = \ln 3 - \ln 4 \; & &: \;\text{Logaritmlag 1 i VL + 2 i HL}\\<br />
\ln\,((x+1) \cdot (x-1)) & = \ln\,\left({3 \over 4}\right) \; & &: \;\text{Konjugatregeln i VL}\\<br />
\ln\,(x^2-1) & = \ln\,\left({3 \over 4}\right) \; & &\;| \;e\,^{\cdot}\\<br />
x^2 - 1 & = {3 \over 4} \\<br />
x^2 & = {3 \over 4} + 1 \\<br />
x^2 & = {7 \over 4} \\<br />
x & = {1 \over 2} \, \sqrt{7}<br />
\end{align}</math></div>
Taifun