Lösung 3.4:1b

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In the equation, both sides are positive because the factors
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<center> [[Image:3_4_1b-1(2).gif]] </center>
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<math>e^{x}</math>
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and
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<math>3^{-x}</math>
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<center> [[Image:3_4_1b-2(2).gif]] </center>
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are positive regardless of the value of
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<math>x</math>
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(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,
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<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>
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Using the log law, we can divide up the products into several logarithmic terms,
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<math>\ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}</math>
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and using the law
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<math>\ln a^{b}=b\centerdot \ln a</math>, we can get rid of
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<math>x</math>
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from the exponents:
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<math>\ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3</math>
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Collecting together
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<math>x</math>
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on one side and the other terms on the other,
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<math>x\ln e+x\ln 3=\ln 2-\ln 13</math>
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Take out
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<math>x</math>
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on the left-hand side and use
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<math>\ln e=1</math>
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:
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<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>
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Then, solve for
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<math>x</math>
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:
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<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>
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NOTE: Because
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<math>\ln 2<\ln 13</math>, we can write the answer as
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<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>
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in order to indicate that
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<math>x</math>
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is negative.

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.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.4:1b“