Lösung 1.3:4b
Aus Online Mathematik Brückenkurs 1
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Version vom 13:35, 22. Okt. 2008
The numbers 9 and 27 can both be written as powers of 3,
| \displaystyle \begin{align}
9 &= 3\cdot 3 = 3^{2}\,,\\[5pt] 27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3}\textrm{.} \end{align} |
Thus, all factors in the expression can be written using a common base and the whole product can be simplified using the power rules
| \displaystyle \begin{align}
3^{13}\cdot 9^{-3}\cdot 27^{-2} &= 3^{13}\cdot (3^{2})^{-3}\cdot (3^{3})^{-2}\\[3pt] &= 3^{13}\cdot 3^{2\cdot (-3)}\cdot 3^{3\cdot (-2)}\\[3pt] &= 3^{13}\cdot 3^{-6}\cdot 3^{-6}\\[3pt] &= 3^{13-6-6}\\[3pt] &= 3^{1}\\[3pt] &= 3\,\textrm{.} \end{align} |
