Lösung 3.4:1a

Aus Online Mathematik Brückenkurs 1

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We solve an equation of this type by taking the natural logarithm of both sides:
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We solve an equation of this type by taking the natural logarithm of both sides
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{{Displayed math||<math>\ln e^x = \ln 13\,\textrm{.}</math>}}
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<math>\ln e^{x}=\ln 13</math>
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Then, using the log law <math>\ln a^{b} = b\cdot \ln a</math>, the unknown <math>x</math> can then be moved down as a factor on the left-hand side
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{{Displayed math||<math>x\cdot \ln e = \ln 13</math>}}
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Then, using the log law
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and then it is simple to solve for <math>x</math>,
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<math>\ln a^{b}=b\centerdot \ln a</math>, the unknown
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<math>x</math>
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can then be moved down as a factor on the left-hand side
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{{Displayed math||<math>x = \frac{\ln 13}{\ln e} = \frac{\ln 13}{1} = \ln 13\,\textrm{.}</math>}}
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<math>x\centerdot \ln e=\ln 13</math>
 
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Note: One thing that we haven't mentioned in this solution is that before we carry out the first step and take the logarithm of both sides of the equation, it is necessary to be certain that the left- and right-hand sides are positive (it is not possible to take the logarithm of a negative number). Because the equation is
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and then it is simple to solve for
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{{Displayed math||<math>e^x=13</math>}}
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<math>x</math>
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,
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we see directly that the right-hand side is positive. The left-hand side is also positive because “a positive number <math>e\approx 2\textrm{.}718</math> raised to a number” always gives a positive number.
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<math>x=\frac{\ln 13}{\ln e}=\frac{\ln 13}{1}=\ln 13</math>
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NOTE: One thing that we haven't mentioned in this solution is that before we carry out the first step and take the logarithm of both sides of the equation, it is necessary to be certain that the left- and right-hand sides are positive (it is not possible to take the logarithm of a negative number). Because the equation is
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<math>e^{x}=13</math>
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we see directly that the right-hand side is positive. The left-hand side is also positive because “a positive number
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<math>e\approx 2.718</math>
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raised to a number” always gives a positive number.
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Version vom 08:45, 2. Okt. 2008

We solve an equation of this type by taking the natural logarithm of both sides

Vorlage:Displayed math

Then, using the log law \displaystyle \ln a^{b} = b\cdot \ln a, the unknown \displaystyle x can then be moved down as a factor on the left-hand side

Vorlage:Displayed math

and then it is simple to solve for \displaystyle x,

Vorlage:Displayed math


Note: One thing that we haven't mentioned in this solution is that before we carry out the first step and take the logarithm of both sides of the equation, it is necessary to be certain that the left- and right-hand sides are positive (it is not possible to take the logarithm of a negative number). Because the equation is

Vorlage:Displayed math

we see directly that the right-hand side is positive. The left-hand side is also positive because “a positive number \displaystyle e\approx 2\textrm{.}718 raised to a number” always gives a positive number.

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