The simplify/ln function is used to simplify logarithmic expressions. It applies the following simplifications whenever it can determine that the appropriate conditions hold:

In the case of an integer argument to ln, the integer is factored and the logarithm is returned as a sum of logarithms.

In the case of a sum of terms as the argument to ln, the integer content of the sum is factored out and the logarithm is returned as a sum of two logarithms.

Making the appropriate assumptions on the names in the expression to be simplified (see assume) provides simplify with enough information to apply the above identities correctly.

$\mathrm{simplify}\left(\ln \left(x^3\right)\,\ln \right)$

$\ln \left(x^3\right)$

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

$\ln \left(xy\right)$

$\mathrm{simplify}\left(\ln \left({\ⅇ}^x\right)\,\ln \right)$

$\ln \left({\ⅇ}^x\right)$

$\mathrm{simplify}\left(\ln \left(y^n\right)\,\ln \right)$

$\ln \left(y^n\right)$

$\mathrm{assume}\left(n\,\mathrm{even}\right)$

$\mathrm{assume}\left(x\,\mathrm{real}\right)$

$\mathrm{assume}\left(y<0\right)$

$\mathrm{simplify}\left(\ln \left(x^3\right)\&comma;\ln \right)$

$\ln \left({\mathrm{x~}}^3\right)$

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

$\ln \left(-\mathrm{y~}\right)\&plus;\ln \left(-\mathrm{x~}\right)$

$\mathrm{simplify}\left(\ln \left({\&ExponentialE;}^x\right)\right)$

$\mathrm{x~}$

$\mathrm{simplify}\left(\ln \left(y^3\right)\&comma;\ln \right)$

$3\ln \left(-\mathrm{y~}\right)\&plus;\mathrm{I}\mathrm{\&pi;}$

$\mathrm{simplify}\left(\ln \left(y^n\right)\&comma;\ln \right)$

$\mathrm{n~}\ln \left(-\mathrm{y~}\right)$

$\mathrm{simplify}\left(\ln \left(-40a\&plus;15b\right)\&comma;\ln \right)$

$\ln \left(5\right)\&plus;\ln \left(-8a\&plus;3b\right)$

$\mathrm{simplify}\left(\ln \left(345366\right)\&comma;\ln \right)$

$\ln \left(2\right)\&plus;2\ln \left(3\right)\&plus;\ln \left(7\right)\&plus;\ln \left(2741\right)$