Aus Online Mathematik Brückenkurs 1

Version vom 09:30, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.4:1b moved to Solution 3.4:1b: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 12:55, 12. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	In the equation, both sides are positive because the factors
-	<center> [[Image:3_4_1b-1(2).gif]] </center>	+	<math>e^{x}</math>
-	{{NAVCONTENT_STOP}}	+	and
-	{{NAVCONTENT_START}}	+	<math>3^{-x}</math>
-	<center> [[Image:3_4_1b-2(2).gif]] </center>	+	are positive regardless of the value of
-	{{NAVCONTENT_STOP}}	+	<math>x</math>
		+	(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,
		+
		+
		+	<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>
		+
		+
		+	Using the log law, we can divide up the products into several logarithmic terms,
		+
		+
		+	<math>\ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}</math>
		+
		+
		+	and using the law
		+	<math>\ln a^{b}=b\centerdot \ln a</math>, we can get rid of
		+	<math>x</math>
		+	from the exponents:
		+
		+
		+	<math>\ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3</math>
		+
		+
		+	Collecting together
		+	<math>x</math>
		+	on one side and the other terms on the other,
		+
		+
		+	<math>x\ln e+x\ln 3=\ln 2-\ln 13</math>
		+
		+
		+	Take out
		+	<math>x</math>
		+	on the left-hand side and use
		+	<math>\ln e=1</math>
		+	:
		+
		+
		+	<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>
		+
		+
		+	Then, solve for
		+	<math>x</math>
		+	:
		+
		+
		+	<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>
		+
		+
		+	NOTE: Because
		+	<math>\ln 2<\ln 13</math>, we can write the answer as
		+
		+
		+	<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>
		+
		+
		+	in order to indicate that
		+	<math>x</math>
		+	is negative.

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Version vom 12:55, 12. Sep. 2008

In the equation, both sides are positive because the factors \displaystyle e^{x} and \displaystyle 3^{-x} are positive regardless of the value of \displaystyle x (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,

\displaystyle \ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)

Using the log law, we can divide up the products into several logarithmic terms,

\displaystyle \ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}

and using the law \displaystyle \ln a^{b}=b\centerdot \ln a, we can get rid of \displaystyle x from the exponents:

\displaystyle \ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3

Collecting together \displaystyle x on one side and the other terms on the other,

\displaystyle x\ln e+x\ln 3=\ln 2-\ln 13

Take out \displaystyle x on the left-hand side and use \displaystyle \ln e=1

\displaystyle x\left( 1+\ln 3 \right)=\ln 2-\ln 13

Then, solve for \displaystyle x

\displaystyle x=\frac{\ln 2-\ln 13}{1+\ln 3}

NOTE: Because \displaystyle \ln 2<\ln 13, we can write the answer as

\displaystyle x=-\frac{\ln 13-\ln 2}{1+\ln 3}

in order to indicate that \displaystyle x is negative.