Aus Online Mathematik Brückenkurs 1

Version vom 10:50, 2. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:45, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 3:		Zeile 3:
	First, we take logs of both sides,		First, we take logs of both sides,
-	{{Displayed math||<math>\ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}</math>}}	+	{{Abgesetzte Formel||<math>\ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}</math>}}
	and use the log laws to make <math>x</math> more accessible,		and use the log laws to make <math>x</math> more accessible,
-	{{Displayed math||<math>\ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>\ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}</math>}}
	Then, collect together the <math>x</math> terms on the left-hand side,		Then, collect together the <math>x</math> terms on the left-hand side,
-	{{Displayed math||<math>x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}</math>}}
	The solution is now		The solution is now
-	{{Displayed math||<math>x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}</math>}}

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Version vom 08:45, 22. Okt. 2008

The equation has the same form as the equation in exercise b and we can therefore use the same strategy.

First, we take logs of both sides,

\displaystyle \ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}

and use the log laws to make \displaystyle x more accessible,

\displaystyle \ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}

Then, collect together the \displaystyle x terms on the left-hand side,

\displaystyle x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}

The solution is now

\displaystyle x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}