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| student:mathematics:logarithms [2018/08/26 14:33] – created bernstdh | student:mathematics:logarithms [2018/08/26 15:08] (current) – bernstdh | ||
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| - | ===== Mathematical Foundations: | + | ===== Logarithms ===== |
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| - \(\log u/v = \log u - \log v\) | - \(\log u/v = \log u - \log v\) | ||
| - \(\log u^{k} = k \log u \mbox{ for } u \gt 0\) | - \(\log u^{k} = k \log u \mbox{ for } u \gt 0\) | ||
| - | - \(\log_{b}u = (\log_{b}a) (\log_{a}u) \mbox{ for } u \gt 0\) | + | - \(\log_{b}u = (\log_{b}a) (\log_{a}u) \mbox{ for } u \gt 0\) \\ To prove this result let \(u= a^{p}\) (so that \(p=\log_{a}u\)). |
| - | To prove this result let \(u= a^{p}\) (so that \(p=\log_{a}u\)). | + | |
| - | \(\log_{b}u = \log_{b}a^{p} = p\log_{b}a = (\log_{b}a) (\log_{a}u)\) | + | |
| - \(\log_{b}u = \frac{1}{\log_{u}b}\) | - \(\log_{b}u = \frac{1}{\log_{u}b}\) | ||
| - | - Combining the previous two rules yields the following " | + | - Combining the previous two rules yields the following " |
| - | \\ | + | - \( \frac{d}{du} \log_{b}u= \frac{1}{( u \ln b)}\) where \(\ln\) denotes the natural logarithm |
| - | \(\log_{b}u = \frac{\log_{a}u}{\log_{a}b} \mbox{ for } u \gt 0\) | + | |
| - | - \( \frac{d}{du} \log_{b}u= \frac{1}{( u \ln b)}\) where \(\ln\) denotes the natural logarithm | + | |