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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,
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<center> [[Bild:3_3_2-f.gif]] </center>
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:*<math>\ \lg (ab) = \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>\lg 10 = 1\,</math>.
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In our case, we have
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{{Displayed math||<math>\lg 10^{3} = 3\cdot \lg 10 = 3\cdot 1 = 3\,\textrm{.}</math>}}

Current revision

Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,

  •  lg(ab)=lga+lgb
  •  lgab=blga

and to simplify expressions first. By working in this way, one only needs, in principle, to learn that lg10=1.

In our case, we have

lg103=3lg10=31=3.
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