Solution 3.3:2f
From Förberedande kurs i matematik 1
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| - | + | Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,  | |
| - | <  | + | |
| - | {{  | + | |
| + | <math>\lg \left( ab \right)=\lg a+\lg b</math>  | ||
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| + | |||
| + | <math>\lg a^{b}=b\lg a</math>  | ||
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| + | and to simplify expressions first. By working in this way, one only needs, in principle, to learn that   | ||
| + | <math>\text{lg 1}0\text{ }=\text{1}</math>.  | ||
| + | |||
| + | In our case, we have  | ||
| + | |||
| + | |||
| + | <math>\lg 10^{3}=3\centerdot \lg 10=3\centerdot 1=3</math>.  | ||
Revision as of 13:22, 25 September 2008
Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,
\displaystyle \lg \left( ab \right)=\lg a+\lg b
\displaystyle \lg a^{b}=b\lg a
and to simplify expressions first. By working in this way, one only needs, in principle, to learn that \displaystyle \text{lg 1}0\text{ }=\text{1}.
In our case, we have
\displaystyle \lg 10^{3}=3\centerdot \lg 10=3\centerdot 1=3.
