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Solution 3.3:2f

From Förberedande kurs i matematik 1

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m (Lösning 3.3:2f moved to Solution 3.3:2f: Robot: moved page)
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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,
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<center> [[Image:3_3_2f.gif]] </center>
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<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
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<math>\text{lg 1}0\text{ }=\text{1}</math>.
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In our case, we have
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<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,


lgab=lga+lgb 


lgab=blga

and to simplify expressions first. By working in this way, one only needs, in principle, to learn that lg 10 =1.

In our case, we have


lg103=3lg10=31=3.

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