Lösung 3.3:3a
Aus Online Mathematik Brückenkurs 1
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By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives | By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}} |
where we have used <math>\log _{2}2 = 1\,</math>. | where we have used <math>\log _{2}2 = 1\,</math>. | ||
Version vom 08:42, 22. Okt. 2008
By writing the argument \displaystyle 8 as \displaystyle 8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3, the logarithm law, \displaystyle \lg a^b = b\lg a, gives
| \displaystyle \log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,, |
where we have used \displaystyle \log _{2}2 = 1\,.
