Lösung 3.3:3d
Aus Online Mathematik Brückenkurs 1
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Version vom 14:34, 22. Okt. 2008
We write the argument of \displaystyle \log_{3} as a power of 3,
| \displaystyle 9\cdot 3^{1/3} = 3^2\cdot 3^{1/3} = 3^{2+1/3} = 3^{7/3}\,, |
and then simplify the expression with the logarithm laws
| \displaystyle \log _3 (9\cdot 3^{1/3}) = \log_3 3^{7/3} = \frac{7}{3}\cdot \log_3 3 = \frac{7}{3}\cdot 1 = \frac{7}{3}\,\textrm{.} |
