Lösung 3.3:2g
Aus Online Mathematik Brückenkurs 1
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by using the log law <math>b\lg a = \lg a^b</math>. This gives | by using the log law <math>b\lg a = \lg a^b</math>. This gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>10^{-\lg 0\textrm{.}1}=10^{\lg 0\textrm{.}1^{-1}}=0\textrm{.}1^{-1}=\frac{1}{0\textrm{.}1}=10\,\textrm{.}</math>}} |
Version vom 08:42, 22. Okt. 2008
We know that \displaystyle 10^{\lg x} = x, so therefore we rewrite the exponent as \displaystyle -\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1} by using the log law \displaystyle b\lg a = \lg a^b. This gives
| \displaystyle 10^{-\lg 0\textrm{.}1}=10^{\lg 0\textrm{.}1^{-1}}=0\textrm{.}1^{-1}=\frac{1}{0\textrm{.}1}=10\,\textrm{.} |
