Solution 3.3:3d
From Förberedande kurs i matematik 1
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| - | {{  | + | We write the argument of   | 
| - | <  | + | <math>\log _{3}</math>  | 
| - | {{  | + | as a power of   | 
| + | <math>\text{3}</math>,  | ||
| + | |||
| + | |||
| + | <math>9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}</math>  | ||
| + | |||
| + | |||
| + | and then simplify the expression with the logarithm laws:  | ||
| + | |||
| + | |||
| + | <math>\log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.</math>  | ||
Revision as of 14:19, 25 September 2008
We write the argument of \displaystyle \log _{3} as a power of \displaystyle \text{3},
\displaystyle 9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}
and then simplify the expression with the logarithm laws:
  
\displaystyle \log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.
