From Förberedande kurs i matematik 1

Revision as of 09:30, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.4:1c moved to Solution 3.4:1c: Robot: moved page) ← Previous diff		Revision as of 13:22, 12 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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-	{{NAVCONTENT_START}}	+	The equation has the same form as the equation in exercise c and we can therefore use the same strategy.
-	<center> [[Image:3_4_1c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	First, we take logs of both sides,
		+
		+
		+	<math>\ln \left( 3e^{x} \right)=\ln \left( 7\centerdot 2^{x} \right)</math>
		+
		+
		+	and use the log laws to make
		+	<math>x</math>
		+	more accessible:
		+
		+
		+	<math>\ln 3+x\centerdot \ln e=\ln 7+x\centerdot \ln 2</math>
		+
		+
		+	Then, collect together the <math>x</math> terms on the left-hand side:
		+
		+
		+	<math>x\left( \ln e-\ln 2 \right)=\ln 7-\ln 3</math>
		+
		+
		+	The solution is now
		+
		+
		+	<math>x=\frac{\ln 7-\ln 3}{\ln e-\ln 2}=\frac{\ln 7-\ln 3}{1-\ln 2}</math>

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Revision as of 13:22, 12 September 2008

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

First, we take logs of both sides,

\displaystyle \ln \left( 3e^{x} \right)=\ln \left( 7\centerdot 2^{x} \right)

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

\displaystyle \ln 3+x\centerdot \ln e=\ln 7+x\centerdot \ln 2

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

\displaystyle x\left( \ln e-\ln 2 \right)=\ln 7-\ln 3

The solution is now

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