Solution 3.3:3a
From Förberedande kurs i matematik 1
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| - | {{  | + | By writing the argument   | 
| - | <  | + | <math>\text{8}</math>  | 
| - | {{  | + | as   | 
| + | <math>8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3}</math>, the logarithm law,   | ||
| + | <math>\lg a^{b}=b\lg a</math>, gives  | ||
| + | |||
| + | |||
| + | <math>\log _{2}8=\log _{2}2^{3}=3\centerdot \log _{2}2=3\centerdot 1=3</math>  | ||
| + | |||
| + | |||
| + | where we have used   | ||
| + | <math>\log _{2}2=1</math>.  | ||
Revision as of 13:57, 25 September 2008
By writing the argument \displaystyle \text{8} as \displaystyle 8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3}, the logarithm law, \displaystyle \lg a^{b}=b\lg a, gives
 
\displaystyle \log _{2}8=\log _{2}2^{3}=3\centerdot \log _{2}2=3\centerdot 1=3
where we have used 
\displaystyle \log _{2}2=1.
