Solution 1.3:1a

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Current revision (12:39, 22 September 2008) (edit) (undo)
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We calculate
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We calculate <math>2^{3}</math> first and then <math>3^{2}</math>,
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<math>2^{3}</math>
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first and then
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<math>3^{2}</math>
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<math>\begin{align}
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& 2^{3}=2\centerdot 2\centerdot 2=4\centerdot 2=8, \\
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& \\
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& 3^{2}=3\centerdot 3=9 \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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2^{3} &= 2\cdot 2\cdot 2 = 4\cdot 2 = 8\,,\\[5pt]
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3^{2} &= 3\cdot 3=9\,\textrm{.}
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\end{align}</math>}}
Hence,
Hence,
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{{Displayed math||<math>2^{3}\cdot 3^{2}=8\cdot 9=72\,</math>.}}
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<math>2^{3}\centerdot 3^{2}=8\centerdot 9=72</math>
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Current revision

We calculate \displaystyle 2^{3} first and then \displaystyle 3^{2},

\displaystyle \begin{align}

2^{3} &= 2\cdot 2\cdot 2 = 4\cdot 2 = 8\,,\\[5pt] 3^{2} &= 3\cdot 3=9\,\textrm{.} \end{align}

Hence,

\displaystyle 2^{3}\cdot 3^{2}=8\cdot 9=72\,.
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