Solution 3.3:2h
From Förberedande kurs i matematik 1
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- | The argument in the logarithm can be rewritten as | + | The argument in the logarithm can be rewritten as <math>\frac{1}{10^{2}} = 10^{-2}</math> and then the log law <math>\lg a^b = b\lg a</math> gives the rest |
- | <math>\frac{1}{10^{2}}=10^{-2}</math> | + | |
- | and then the log law | + | |
- | <math>\lg a^ | + | |
- | gives the rest | + | |
- | + | {{Displayed math||<math>\lg \frac{1}{10^2} = \lg 10^{-2} = (-2)\cdot \lg 10 = (-2)\cdot 1 = -2\,\textrm{.}</math>}} | |
- | <math>\lg \frac{1}{10^ | + |
Current revision
The argument in the logarithm can be rewritten as \displaystyle \frac{1}{10^{2}} = 10^{-2} and then the log law \displaystyle \lg a^b = b\lg a gives the rest
\displaystyle \lg \frac{1}{10^2} = \lg 10^{-2} = (-2)\cdot \lg 10 = (-2)\cdot 1 = -2\,\textrm{.} |