Solution 3.3:3b
From Förberedande kurs i matematik 1
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- | { | + | Because we are working with <math>\log _{9}</math>, we express 1/3 as a power of 9, |
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- | {{ | + | {{Displayed math||<math>\frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}</math>}} |
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+ | Using the logarithm laws, we get | ||
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+ | {{Displayed math||<math>\log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}</math>}} |
Current revision
Because we are working with \displaystyle \log _{9}, we express 1/3 as a power of 9,
\displaystyle \frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.} |
Using the logarithm laws, we get
\displaystyle \log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.} |