Lösung 3.4:1b
Aus Online Mathematik Brückenkurs 1
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In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides, | In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}} |
Using the log laws, we can divide up the products into several logarithmic terms, | Using the log laws, we can divide up the products into several logarithmic terms, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}} |
and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents | and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}} |
Collect <math>x</math> on one side and the other terms on the other, | Collect <math>x</math> on one side and the other terms on the other, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}} |
Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>, | Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}} |
Then, solve for <math>x</math>, | Then, solve for <math>x</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}} |
Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as | Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}} |
to indicate that <math>x</math> is negative. | to indicate that <math>x</math> is negative. | ||
Version vom 08:45, 22. Okt. 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 sides,
| \displaystyle \ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.} |
Using the log laws, 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\cdot \ln a, we can get rid of \displaystyle x from the exponents
| \displaystyle \ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.} |
Collect \displaystyle x on one side and the other terms on the other,
| \displaystyle x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.} |
Take out \displaystyle x on the left-hand side and use \displaystyle \ln e=1,
| \displaystyle x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.} |
Then, solve for \displaystyle x,
| \displaystyle x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.} |
Note: Because \displaystyle \ln 2 < \ln 13, we can write the answer as
| \displaystyle x=-\frac{\ln 13-\ln 2}{1+\ln 3} |
to indicate that \displaystyle x is negative.
