如何简化Maxima CAS中的log(8)/log(2)？

Question

我想简化 log(8)/log(2)

我知道

log(8)/log(2) = log(2^3)/log(2) = 3*log(2)/log(2) = 3

这在 Maxima 中是可能的，但对我不起作用：

Maxima 5.41.0 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.12
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) log(8)/log(2);
                                    log(8)
(%o1)                               ------
                                    log(2)
(%i2) logexpand;
(%o2)                                true
(%i3) log(2^3)/log(2);
(%o3)                               log(8)
                                    ------
                                    log(2)

(%i4) logexpand;
(%o4)                                true

我用：

round(float(log(8)/log(2));

但我认为这不是最好的解决方案（我使用整数）

问题：

1. 怎么做 ？
2. 为什么它在 Maxima doc 中有效，但在我的 Maxima 中无效？

Answer 1

这在 Maxima 5.43.0 中适用于我：

(%i1) radcan(log(8)/log(2));
(%o1)                                  3
(%i2) radcan(log(2^3)/log(2));
(%o2)                                  3

马克西玛说

 -- Function: radcan (<expr>)

     Simplifies <expr>, which can contain logs, exponentials, and
     radicals, by converting it into a form which is canonical over a
     large class of expressions and a given ordering of variables; that
     is, all functionally equivalent forms are mapped into a unique
     form.  For a somewhat larger class of expressions, 'radcan'
     produces a regular form.  Two equivalent expressions in this class
     do not necessarily have the same appearance, but their difference
     can be simplified by 'radcan' to zero.

在这种情况下，它对数字进行因式分解8，然后将 3 的幂移出对数，从而消除 2 的剩余对数：

(%i3) radcan(log(8));
(%o3)                              3 log(2)