Solution 1.3:4c

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Current revision (14:05, 22 September 2008) (edit) (undo)
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The whole expression consists of factors having a base of
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The whole expression consists of factors having a base of 5 so the power rules can be used to simplify the expression first
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<math>5</math>;
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so the power rules can be use to
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{{Displayed math||<math>\begin{align}
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simplify the expression first:
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\frac{5^{12}}{5^{-4}}\cdot \bigl( 5^{2} \bigr)^{-6}
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&= \frac{5^{12}}{5^{-4}}\cdot 5^{2\cdot (-6)}\\[3pt]
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&= \frac{5^{12}}{5^{-4}}\cdot 5^{-12}\\[3pt]
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<math>\begin{align}
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&= \frac{5^{12}\cdot 5^{-12}}{5^{-4}}\\[3pt]
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& \frac{5^{12}}{5^{-4}}\centerdot \left( 5^{2} \right)^{-6}=\frac{5^{12}}{5^{-4}}\centerdot 5^{2\centerdot \left( -6 \right)}=\frac{5^{12}}{5^{-4}}\centerdot 5^{-12}=\frac{5^{12}\centerdot 5^{-12}}{5^{-4}} \\
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&= \frac{5^{12-12}}{5^{-4}}\\[3pt]
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& \\
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&= \frac{5^{0}}{5^{-4}}\\[3pt]
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& =\frac{5^{12-12}}{5^{-4}}=\frac{5^{0}}{5^{-4}}=5^{0-\left( -4 \right)}=5^{4}=5\centerdot 5\centerdot 5\centerdot 5=625 \\
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&= 5^{0-(-4)}\\[3pt]
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\end{align}</math>
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&= 5^{4}\\[3pt]
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&= 5\cdot 5\cdot 5\cdot 5\\[3pt]
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&= 625\,\textrm{.}
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\end{align}</math>}}

Current revision

The whole expression consists of factors having a base of 5 so the power rules can be used to simplify the expression first

\displaystyle \begin{align}

\frac{5^{12}}{5^{-4}}\cdot \bigl( 5^{2} \bigr)^{-6} &= \frac{5^{12}}{5^{-4}}\cdot 5^{2\cdot (-6)}\\[3pt] &= \frac{5^{12}}{5^{-4}}\cdot 5^{-12}\\[3pt] &= \frac{5^{12}\cdot 5^{-12}}{5^{-4}}\\[3pt] &= \frac{5^{12-12}}{5^{-4}}\\[3pt] &= \frac{5^{0}}{5^{-4}}\\[3pt] &= 5^{0-(-4)}\\[3pt] &= 5^{4}\\[3pt] &= 5\cdot 5\cdot 5\cdot 5\\[3pt] &= 625\,\textrm{.} \end{align}

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