Lösung 3.3:3h
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
| Zeile 1: | Zeile 1: | ||
Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that | Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}} |
where we have used that <math>\log_{a}a = 1\,</math>. | where we have used that <math>\log_{a}a = 1\,</math>. | ||
Version vom 08:43, 22. Okt. 2008
Because \displaystyle a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}, the logarithm law, \displaystyle b\lg a = \lg a^b, gives that
| \displaystyle \log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,, |
where we have used that \displaystyle \log_{a}a = 1\,.
Note: In this exercise, we assume, implicitly, that \displaystyle a > 0 and \displaystyle a\ne 1\,.
