Why is this like it is? :D $$\dfrac{1}{\log_ae} = \ln(a)$$ I'm solving some exercises and I ran up to this? Maybe it's really banal, but please explain me...
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1$\ln a \cdot \log_a e = \log_a e^{\ln a} = \log_a a = 1$. – njguliyev Oct 27 '13 at 15:52
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Note that $\ln a = \log_e a$. $$\log_a e = \frac{\ln e}{\ln a} = \frac 1{\ln a}$$
So, $\dfrac{1}{\log_a e} = \cdots $
amWhy
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We can prove $$\log_ab=\frac{\log_cb}{\log_ca}$$ where $a,c >0,\ne1$
$$\implies \log_ab\cdot\log_ba=\cdots=1$$
and conventionally Natural logarithm is written as $\displaystyle \ln a$ which means $\displaystyle \log_ea$
lab bhattacharjee
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You've got the change of base wrong; you've inverted the relationship, which should be $$\log_a b = \dfrac{\log_c b}{\log_c a}$$ – amWhy Oct 27 '13 at 16:03
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