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Solution 3.3:2g

From Förberedande kurs i matematik 1

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Current revision (06:20, 2 October 2008) (edit) (undo)
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We know that
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We know that <math>10^{\lg x} = x</math>, so therefore we rewrite the exponent as
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<math>10^{\lg x}=x</math>, so therefore we rewrite the exponent as
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<math>-\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1}</math>
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<math>-\lg 0.1=\left( -1 \right)\centerdot \lg 0.1=\lg 0.1^{-1}</math>
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by using the log law <math>b\lg a = \lg a^b</math>. This gives
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by using the log law
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<math>b\lg a=\lg a^{b}</math>. This gives
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{{Displayed math||<math>10^{-\lg 0\textrm{.}1}=10^{\lg 0\textrm{.}1^{-1}}=0\textrm{.}1^{-1}=\frac{1}{0\textrm{.}1}=10\,\textrm{.}</math>}}
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<math>10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10</math>
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Current revision

We know that 10lgx=x, so therefore we rewrite the exponent as −lg0.1=(−1)lg0.1=lg0.1−1 by using the log law blga=lgab. This gives

10−lg0.1=10lg0.1−1=0.1−1=10.1=10.
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