Solution 1.3:6f

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Current revision (07:33, 23 September 2008) (edit) (undo)
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We can factorize the exponents
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We can factorize the exponents 40 and 56 as
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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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{{Displayed math||<math>\begin{align}
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40 &= 4\cdot 10 = 2\cdot 2\cdot 2\cdot 5 = 2^{3}\cdot 5 \\[3pt]
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56 &= 7\cdot 8 = 7\cdot 2\cdot 4 = 7\cdot 2\cdot 2\cdot 2 = 2^{3}\cdot 7
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\end{align}</math>}}
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<math>\begin{align}
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and we then see that they have <math>2^{3} = 8</math> as a common factor. We can take this factor out as an "outer" exponent
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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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{{Displayed math||<math>\begin{align}
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3^{40} &= 3^{5\cdot 8} = \bigl(3^{5}\bigr)^{8} = (3\cdot 3\cdot 3\cdot 3\cdot 3)^{8} = 243^{8}\,,\\[3pt]
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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{.}
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\end{align}</math>}}
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and we then see that they have
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This shows that <math>3^{40} = 243^{8}</math> is bigger than <math>2^{56} = 128^{8}</math>.
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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>
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Current revision

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}.

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