Lösung 1.3:6f
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Version vom 13:40, 22. Okt. 2008
We can factorize the exponents 40 and 56 as
| \displaystyle \begin{align}
40 &= 4\cdot 10 = 2\cdot 2\cdot 2\cdot 5 = 2^{3}\cdot 5 \\[3pt] 56 &= 7\cdot 8 = 7\cdot 2\cdot 4 = 7\cdot 2\cdot 2\cdot 2 = 2^{3}\cdot 7 \end{align} |
and we then see that they have \displaystyle 2^{3} = 8 as a common factor. We can take this factor out as an "outer" exponent
| \displaystyle \begin{align}
3^{40} &= 3^{5\cdot 8} = \bigl(3^{5}\bigr)^{8} = (3\cdot 3\cdot 3\cdot 3\cdot 3)^{8} = 243^{8}\,,\\[3pt] 2^{56} &= 2^{7\cdot 8} = \bigl(2^{7}\bigr)^{8} = (2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2)^{8} = 128^{8}\,\textrm{.} \end{align} |
This shows that \displaystyle 3^{40} = 243^{8} is bigger than \displaystyle 2^{56} = 128^{8}.
