Lösung 1.3:6f

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We can factorize the exponents
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<center> [[Image:1_3_6f.gif]] </center>
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<math>40</math>
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and
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<math>56</math>
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as
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<math>\begin{align}
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& 40=4\centerdot 10=2\centerdot 2\centerdot 2\centerdot 5=2^{3}\centerdot 5 \\
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& \\
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& 56=7\centerdot 8=7\centerdot 2\centerdot 4=7\centerdot 2\centerdot 2\centerdot 2=2^{3}\centerdot 7 \\
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\end{align}</math>
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and we then see that they have
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<math>2^{3}=8</math>
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as a common factor. We can take this factor out as an "outer" exponent:
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<math>\begin{align}
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& 3^{40}=3^{5\centerdot 8}=\left( 3^{5} \right)^{8}=\left( 3\centerdot 3\centerdot 3\centerdot 3\centerdot 3 \right)^{8}=243^{8} \\
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& \\
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& 2^{56}=2^{7\centerdot 8}=\left( 2^{7} \right)^{8}=\left( 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2 \right)^{8}=128^{8} \\
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\end{align}</math>
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This shows that
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<math>3^{40}=243^{8}</math>
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is bigger than
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<math>2^{56}=128^{8}</math>

Version vom 13:01, 15. Sep. 2008

We can factorize the exponents \displaystyle 40 and \displaystyle 56 as


\displaystyle \begin{align} & 40=4\centerdot 10=2\centerdot 2\centerdot 2\centerdot 5=2^{3}\centerdot 5 \\ & \\ & 56=7\centerdot 8=7\centerdot 2\centerdot 4=7\centerdot 2\centerdot 2\centerdot 2=2^{3}\centerdot 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\centerdot 8}=\left( 3^{5} \right)^{8}=\left( 3\centerdot 3\centerdot 3\centerdot 3\centerdot 3 \right)^{8}=243^{8} \\ & \\ & 2^{56}=2^{7\centerdot 8}=\left( 2^{7} \right)^{8}=\left( 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2 \right)^{8}=128^{8} \\ \end{align}


This shows that \displaystyle 3^{40}=243^{8} is bigger than \displaystyle 2^{56}=128^{8}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:6f“