Lösung 3.1:8d

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.1:8d moved to Solution 3.1:8d: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
In power form, the expressions become
-
<center> [[Image:3_1_8d-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
<math>\begin{align}
-
<center> [[Image:3_1_8d-2(2).gif]] </center>
+
& \sqrt{2}\left( \sqrt[4]{3} \right)^{3}=2^{{1}/{2}\;}\left( 3^{{1}/{4}\;} \right)^{3}=2^{{1}/{2}\;}3^{{3}/{4}\;}, \\
-
{{NAVCONTENT_STOP}}
+
& \sqrt[3]{2}\centerdot 3=2^{{1}/{3}\;}3^{1} \\
 +
\end{align}</math>
 +
 +
 
 +
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
 +
<math>\frac{1}{2},\ \ \frac{3}{4},\ \ \frac{1}{3}</math>
 +
and
 +
<math>\text{1}</math>
 +
have
 +
<math>\text{3}\centerdot \text{4}=\text{12 }</math>
 +
as the lowest common denominator which we can take out:
 +
 
 +
 
 +
<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
 +
 
 +
 
 +
<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>\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>.

Version vom 11:39, 23. Sep. 2008

In power form, the expressions become


\displaystyle \begin{align} & \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}


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 \displaystyle \frac{1}{2},\ \ \frac{3}{4},\ \ \frac{1}{3} and \displaystyle \text{1} have \displaystyle \text{3}\centerdot \text{4}=\text{12 } as the lowest common denominator which we can take out:


\displaystyle \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}

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

Because


\displaystyle \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

the denominator \displaystyle \text{2}^{\text{4}}\centerdot \text{3}^{\text{12}} is larger than the numerator \displaystyle \text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}, which means that \displaystyle \sqrt[3]{2}\centerdot 3 is larger than \displaystyle \sqrt{2}\left( \sqrt[4]{3} \right)^{3}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:8d“