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 (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,
- \displaystyle \ \lg (ab) = \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 \lg 10 = 1\,.
In our case, we have
| \displaystyle \lg 10^{3} = 3\cdot \lg 10 = 3\cdot 1 = 3\,\textrm{.} |
