The natural logarithm, ln, is the logarithm with base $\&ExponentialE;\&equals;2.71828$...  For $0<x$ we have $\ln \left(x\right)\&equals;y$ <==> $x\&equals;{\&ExponentialE;}^y$.

For complex-valued expressions x, $\ln \left(x\right)\&equals;\ln \left(\left|x\right|\right)\&plus;\mathrm{I}\mathrm{argument}\left(x\right)$, where $-\mathrm{\pi }$<argument(x)<=$\mathrm{\pi }$.  Throughout Maple, this computation is taken to be the definition of the principal branch of the logarithm.

The log function is the general logarithm.  For $0<x$ and $0<b$ we have $\log \&lsqb;b\&rsqb;\left(x\right)\&equals;y$<==>$x\&equals;b^y$.  log is extended to general complex b and x by $\log \&lsqb;b\&rsqb;\left(x\right)\&equals;\frac{\ln \left(x\right)}{\ln \left(b\right)}$.

The default value of the base b is $\&ExponentialE;$.

You can enter  the function log with base b using either the 1-D or 2-D calling sequence.  Similarly, $e$ can also be entered as exp(1) in 1-D.  See exp for more about the exponential function.

$\mathrm{log10}\left(x\right)\&equals;\log \&lsqb;10\&rsqb;\left(x\right)$.

$\ln \left(x\right)\&equals;\log \&lsqb;\&ExponentialE;\&rsqb;\left(x\right)$.

$\ln \left(1\right)$

$0$

$\frac{\&DifferentialD;}{\&DifferentialD;x}\ln \left(x\right)$

$\frac{1}{x}$

$\ln \left(3.14\&plus;2.71I\right)$

$1.422562238\&plus;0.7120258406\mathrm{I}$

$\ln \left(3\&plus;4I\right)$

$\ln \left(3\&plus;4\mathrm{I}\right)$

$\mathrm{evalc}\left(\right)$

$\ln \left(5\right)\&plus;\mathrm{I}\arctan \left(\frac{4}{3}\right)$

$\ln \left(10000\right)$

$4\ln \left(10\right)$

$\log \left(10000\right)$

$4\ln \left(10\right)$

$\log \left({\&ExponentialE;}^3\right)$

$3$

$\mathrm{log10}\left(10000\right)$

$4$

$\log \&lsqb;\&ExponentialE;\&rsqb;\left(x\right)$

$\ln \left(x\right)$

$\log \&lsqb;b\&rsqb;\left(x\right)$

$\frac{\ln \left(x\right)}{\ln \left(b\right)}$

$\mathrm{log10}\left(65\right)$

$\frac{\ln \left(65\right)}{\ln \left(10\right)}$

$\log \&lsqb;10\&rsqb;\left(100\right)$

$2$

$\log \&lsqb;2\&rsqb;\left(\&ExponentialE;\right)$

$\frac{1}{\ln \left(2\right)}$

$\mathrm{evalf}\left(\right)$

$1.442695041$

$\log \&lsqb;5\&rsqb;\left(5x\right)-\log \&lsqb;5\&rsqb;\left(x\right)$

$\frac{\ln \left(5x\right)}{\ln \left(5\right)}-\frac{\ln \left(x\right)}{\ln \left(5\right)}$

$\mathrm{simplify}\left(\right)$

$1$

$\mathrm{solve}\left(\log \&lsqb;6\&rsqb;\left(2y\right)\&equals;2\&comma;y\right)$

$18$

$\mathrm{convert}\left(\arcsin \left(x\right)\&comma;\ln \right)$

$-\mathrm{I}\ln \left(\mathrm{I}x\&plus;\sqrt{-x^2\&plus;1}\right)$