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--- student:mathematics:logarithms [2018/08/26 14:33] – bernstdh
+++ student:mathematics:logarithms [2018/08/26 15:08] (current) – bernstdh
@@ Line 1: / Line 1: @@
-===== Mathematical Foundations: Logarithms =====
+===== Logarithms =====
@@ Line 27: / Line 27: @@
   -  $\log_{b}u = (\log_{b}a) (\log_{a}u) \mbox{ for }  u \gt 0$  \\ To prove this result let   $u= a^{p}$  (so that   $p=\log_{a}u$ ).  Then: \\  $\log_{b}u = \log_{b}a^{p} = p\log_{b}a =  (\log_{b}a) (\log_{a}u)$
   -  $\log_{b}u = \frac{1}{\log_{u}b}$
-  - 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.,  $\log_{e}$ ).

student/mathematics/logarithms.1535308427.txt.gz · Last modified: 2018/08/26 14:33 by bernstdh

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