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student:mathematics:logarithms [2018/08/26 14:33] – bernstdhstudent:mathematics:logarithms [2018/08/26 14:34] – bernstdh
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   - \(\log_{b}u = (\log_{b}a) (\log_{a}u) \mbox{ for }  u \gt 0\) \\ To prove this result let  \(u= a^{p}\) (so that  \(p=\log_{a}u\)).  Then: \\ \(\log_{b}u = \log_{b}a^{p} = p\log_{b}a =  (\log_{b}a) (\log_{a}u)\)   - \(\log_{b}u = (\log_{b}a) (\log_{a}u) \mbox{ for }  u \gt 0\) \\ To prove this result let  \(u= a^{p}\) (so that  \(p=\log_{a}u\)).  Then: \\ \(\log_{b}u = \log_{b}a^{p} = p\log_{b}a =  (\log_{b}a) (\log_{a}u)\)
   - \(\log_{b}u = \frac{1}{\log_{u}b}\)   - \(\log_{b}u = \frac{1}{\log_{u}b}\)
-  - Combining the previous two rules yields the following "change of base rule": +  - Combining the previous two rules yields the following "change of base rule":\\ \(\log_{b}u = \frac{\log_{a}u}{\log_{a}b} \mbox{ for }  u \gt 0\) 
-\\ +  - \( \frac{d}{du} \log_{b}u= \frac{1}{( u \ln b)}\) where \(\ln\) denotes the natural logarithm  (i.e., \(\log_{e}\)).
-\(\log_{b}u = \frac{\log_{a}u}{\log_{a}b} \mbox{ for }  u \gt 0\) +
-  - \( \frac{d}{du} \log_{b}u= \frac{1}{( u \ln b)}\) where \(\ln\) denotes the natural logarithm  i.e., \(\log_{e}\)).+