Solution 3.1:4b
From Förberedande kurs i matematik 1
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| - | {{  | + | By writing   | 
| - | <  | + | <math>0.0\text{27 }</math>  | 
| - | {{  | + | as  | 
| + | <math>\text{27}\cdot \text{1}0^{-\text{3}}</math>, where   | ||
| + | <math>\text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}}</math>  | ||
| + | and   | ||
| + | <math>10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3}</math>  | ||
| + | we see that  | ||
| + | |||
| + | |||
| + | <math>\begin{align}  | ||
| + | & \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\   | ||
| + | & =3\centerdot 0.1=0.3 \\   | ||
| + | \end{align}</math>  | ||
| + | |||
| + | |||
| + | where we have used   | ||
| + | <math>\sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a</math>  | ||
Revision as of 13:58, 22 September 2008
By writing \displaystyle 0.0\text{27 } as \displaystyle \text{27}\cdot \text{1}0^{-\text{3}}, where \displaystyle \text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}} and \displaystyle 10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3} we see that
\displaystyle \begin{align}
& \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\ 
& =3\centerdot 0.1=0.3 \\ 
\end{align}
where we have used 
\displaystyle \sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a
