$(\log n)'=1/n$? Shouldn't it be $(1/n)(1/\ln 10)$? why i see some people write $(\log n)' = \dfrac{1}{n}$? Shouldn't it be $\dfrac{1}{n}\dfrac{1}{\ln 10}$? 
for example,
$$
\begin{array}{rcl}
  \lim\limits_{n \to \infty} \frac{(\log n)^3+3(\log n)^2}{n^{1/2}}
  &=& \dfrac{6(\log n)^2+12(\log n)}{n^{1/2}} \\
  &=& \dfrac{24\log n+24}{n^{1/2}} \\
  &=& \dfrac{48}{n^{1/2}} \\
  &=& 0
\end{array}
$$
Could anyone help to explain? I am new in learning the calculus.
Thanks.
 A: As is noted in the comments, some sources denote the logarithm with a base of $e$ as $\ln$, while others denote the same as $\log=\log_e$. It depends on the context. I know my calculus teacher always uses $\ln$ for the natural logarithm and $\log=\log_{10}$. 
As for your example: 
$\lim\limits_{n \to \infty} \frac{(\log n)^3+3(\log n)^2}{n^{1/2}}$
$\lim\limits_{n \to \infty} \frac{(\log n)^3+3(\log n)^2}{n^{1/2}}=\lim\limits_{n \to \infty} \frac{(\log n)^3}{n^{1/2}}+\lim\limits_{n \to \infty}\frac{3(\log n)^2}{n^{1/2}}$. 
Both of our new limits are of an indeterminate form. Thus, we can apply L'Hopital's rule. 
$\lim\limits_{n \to \infty} \frac{(\log n)^3}{n^{1/2}}+\lim\limits_{n \to \infty}\frac{3(\log n)^2}{n^{1/2}}=
6\lim\limits_{n \to \infty} \frac{(\log n)^2}{n^{1/2}}+12\lim\limits_{n \to \infty}\frac{(\log n)}{n^{1/2}}$. This is still indeterminate. Another application of L'Hopital gives us
$24\lim\limits_{n \to \infty} \frac{(\log n)}{n^{1/2}}+24\lim\limits_{n \to \infty}\frac{1}{n^{1/2}}$. At this point, the second limit is equal to zero. The first limit is unfortunately still indeterminate. Once more, we use L'Hopital. 
Now we have $48\lim\limits_{n \to \infty} \frac{1}{n^{1/2}}$. This also goes to zero.
Thus, 

$$\lim\limits_{n \to \infty} \frac{(\log n)^3+3(\log n)^2}{n^{1/2}}=0$$

