Solution 3.1:8d

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In power form, the expressions become
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<center> [[Image:3_1_8d-1(2).gif]] </center>
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{{Displayed math||<math>\begin{align}
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\sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}
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<center> [[Image:3_1_8d-2(2).gif]] </center>
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&= 2^{1/2}\bigl(3^{1/4}\bigr)^{3}
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= 2^{1/2}3^{3/4},\\[5pt]
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\sqrt[3]{2}\cdot 3
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&= 2^{1/3}3^{1}\,\textrm{.}
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\end{align}</math>}}
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Admittedly, it is true that <math>2^{1/2} > 2^{1/3}</math> and <math>3^1 > 3^{3/4}</math>, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents 1/2, 3/4, 1/3 and 1 have <math>3\cdot 4 = 12</math> as the lowest common denominator which we can take out
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{{Displayed math||<math>\begin{align}
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2^{1/2}3^{3/4}
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&= 2^{6/12}3^{(3\cdot 3)/12}
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= \bigl(2^{6}\cdot 3^{9}\bigr)^{1/12},\\[5pt]
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2^{1/3}3^{1}
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&= 2^{4/12}3^{12/12}
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= \bigl(2^{4}\cdot 3^{12}\bigr)^{1/12}\,\textrm{.}
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\end{align}</math>}}
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Now, we can compare the bases <math>2^6\cdot 3^9</math> and <math>2^4\cdot 3^{12}</math> with each other and so decide which number is larger.
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Because
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{{Displayed math||<math>\frac{2^6\cdot 3^9}{2^4\cdot 3^{12}} = 2^{6-4}3^{9-12} = 2^{2}3^{-3} = \frac{2^{2}}{3^{3}} = \frac{4}{27} < 1</math>}}
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the denominator <math>2^{4}\cdot 3^{12}</math> is larger than the numerator
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<math>2^6\cdot 3^9</math>, which means that <math>\sqrt[3]{2}\cdot 3</math>
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is larger than <math>\sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}</math>.

Current revision

In power form, the expressions become

\displaystyle \begin{align}

\sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3} &= 2^{1/2}\bigl(3^{1/4}\bigr)^{3} = 2^{1/2}3^{3/4},\\[5pt] \sqrt[3]{2}\cdot 3 &= 2^{1/3}3^{1}\,\textrm{.} \end{align}

Admittedly, it is true that \displaystyle 2^{1/2} > 2^{1/3} and \displaystyle 3^1 > 3^{3/4}, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents 1/2, 3/4, 1/3 and 1 have \displaystyle 3\cdot 4 = 12 as the lowest common denominator which we can take out

\displaystyle \begin{align}

2^{1/2}3^{3/4} &= 2^{6/12}3^{(3\cdot 3)/12} = \bigl(2^{6}\cdot 3^{9}\bigr)^{1/12},\\[5pt] 2^{1/3}3^{1} &= 2^{4/12}3^{12/12} = \bigl(2^{4}\cdot 3^{12}\bigr)^{1/12}\,\textrm{.} \end{align}

Now, we can compare the bases \displaystyle 2^6\cdot 3^9 and \displaystyle 2^4\cdot 3^{12} with each other and so decide which number is larger.

Because

\displaystyle \frac{2^6\cdot 3^9}{2^4\cdot 3^{12}} = 2^{6-4}3^{9-12} = 2^{2}3^{-3} = \frac{2^{2}}{3^{3}} = \frac{4}{27} < 1

the denominator \displaystyle 2^{4}\cdot 3^{12} is larger than the numerator \displaystyle 2^6\cdot 3^9, which means that \displaystyle \sqrt[3]{2}\cdot 3 is larger than \displaystyle \sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}.

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