Solution 3.1:8d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:41, 30 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
In power form, the expressions become
In power form, the expressions become
 +
{{Displayed math||<math>\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}</math>}}
-
<math>\begin{align}
+
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
-
& \sqrt{2}\left( \sqrt[4]{3} \right)^{3}=2^{{1}/{2}\;}\left( 3^{{1}/{4}\;} \right)^{3}=2^{{1}/{2}\;}3^{{3}/{4}\;}, \\
+
-
& \sqrt[3]{2}\centerdot 3=2^{{1}/{3}\;}3^{1} \\
+
-
\end{align}</math>
+
-
+
-
Admittedly, it is true that
+
{{Displayed math||<math>\begin{align}
-
<math>2^{{1}/{2}\;}>2^{{1}/{3}\;}</math>
+
2^{1/2}3^{3/4}
-
and
+
&= 2^{6/12}3^{(3\cdot 3)/12}
-
<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
+
= \bigl(2^{6}\cdot 3^{9}\bigr)^{1/12},\\[5pt]
-
<math>\frac{1}{2},\ \ \frac{3}{4},\ \ \frac{1}{3}</math>
+
2^{1/3}3^{1}
-
and
+
&= 2^{4/12}3^{12/12}
-
<math>\text{1}</math>
+
= \bigl(2^{4}\cdot 3^{12}\bigr)^{1/12}\,\textrm{.}
-
have
+
\end{align}</math>}}
-
<math>\text{3}\centerdot \text{4}=\text{12 }</math>
+
-
as the lowest common denominator which we can take out:
+
-
 
+
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.
-
<math>\begin{align}
+
-
& 2^{\frac{1}{2}}3^{\frac{3}{4}}=2^{\frac{6}{12}}3^{\frac{3\centerdot 3}{12}}=\left( 2^{6}\centerdot 3^{9} \right)^{\frac{1}{12}}, \\
+
-
& 2^{\frac{1}{3}}3^{1}=2^{\frac{4}{12}}3^{\frac{12}{12}}=\left( 2^{4}\centerdot 3^{12} \right)^{\frac{1}{12}}. \\
+
-
\end{align}</math>
+
-
 
+
-
Now, we can compare the bases
+
-
<math>\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}</math>
+
-
and
+
-
<math>\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}</math>
+
-
with each other and so decide which number is larger.
+
Because
Because
 +
{{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>}}
-
<math>\frac{\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}}{\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}}=2^{6-4}3^{9-12}=2^{2}3^{-3}=\frac{2^{2}}{3^{3}}=\frac{4}{27}<1</math>
+
the denominator <math>2^{4}\cdot 3^{12}</math> is larger than the numerator
-
 
+
<math>2^6\cdot 3^9</math>, which means that <math>\sqrt[3]{2}\cdot 3</math>
-
the denominator
+
is larger than <math>\sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}</math>.
-
<math>\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}</math>
+
-
is larger than the numerator
+
-
<math>\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}</math>, which means that
+
-
<math>\sqrt[3]{2}\centerdot 3</math>
+
-
is larger than
+
-
<math>\sqrt{2}\left( \sqrt[4]{3} \right)^{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}.

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_3.1:8d"