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I found in some integral equations where they use $\log(n)$ and in some other with $\ln(n)$.

Suppose

$$ \int_{n_0}^{\large\frac{n_0}{2}} \frac{1}{n}dn $$

Which formula should I use ?

$$ \log(n)\ \mbox{or}\ \ln(n) $$

Aditya Hase
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Vinayak
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  • @Amzoti Sorry I couldnt get u.. Which formula should I use ? – Vinayak Nov 22 '14 at 07:11
  • Thank you. And let me know why do we do so ? – Vinayak Nov 22 '14 at 07:15
  • I didn't get anything...Could u pls explain ? – Vinayak Nov 22 '14 at 07:17
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    People in Engineering, and the physical sciences, still often use $\log$ to mean logarithm to the base $10$. In mathematics, $\log$ usually means logarithm to the base $e$, that is, the natural logarithm, often called $\ln$. In mathematics formulas, usually $\log$ and $\ln$ mean the same thing. – André Nicolas Nov 22 '14 at 07:50
  • @Integrator i am sorry to say that it didn't. I am searching why should I use $In$ – Vinayak Nov 22 '14 at 07:54

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It depends who you talk to. For example, a mathematician would say that $$ \log x = \log_e x = \ln x$$ But a computer scientist would sat that $$ \log x = \log_{10} x $$ I avoid these issues by writing $\ln x$ for the natural logarithm. I also explicitly write the base in all other situations, for example $\log_a x$. As for your integral, here are the steps $$ \int_{n_0}^{\frac{n_0}{2}} \frac{1}{n}dn = \ln \left|\frac{n_0}{2}\right|-\ln |n_0|=\ln \left|n_0\right|-\ln |2|-\ln |n_0|=-\ln 2 $$

k170
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