Lösung 3.3:5b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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By using the logarithm laws, | By using the logarithm laws, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt] | \ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt] | ||
\ln a - \ln b &= \ln\frac{a}{b}\,, | \ln a - \ln b &= \ln\frac{a}{b}\,, | ||
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we can collect together the terms into one logarithmic expression | we can collect together the terms into one logarithmic expression | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt] | \ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt] | ||
&= \ln 8 - \ln(4\cdot 2)\\[5pt] | &= \ln 8 - \ln(4\cdot 2)\\[5pt] | ||
Version vom 08:44, 22. Okt. 2008
By using the logarithm laws,
| \displaystyle \begin{align}
\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt] \ln a - \ln b &= \ln\frac{a}{b}\,, \end{align} |
we can collect together the terms into one logarithmic expression
| \displaystyle \begin{align}
\ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt] &= \ln 8 - \ln(4\cdot 2)\\[5pt] &= \ln\frac{8}{4\cdot 2}\\[5pt] &= \ln 1\\[5pt] &= 0\,, \end{align} |
where \displaystyle \ln 1 = 0, since \displaystyle e^{0}=1 (the equality \displaystyle a^{0}=1 holds for all \displaystyle a\ne 0).
