How do you solve ln(-x)=x?

I know that the solution is -0'567... But I don't know what is the process. Thanks.

1 Answer
Jun 6, 2018

See below.

Explanation:

This equation can't be "explicitly" solved, i.e. there's no method leading to an analytic answer.

The best you can do is to approximate the solution, with one of the methods available. For example, you might use bisection.

It works like this: if we want to find the root of a given continuous function #f(x)# and we have an interval #[a,b]# such that

#f(a)f(b)<0#

(i.e., #f(a)# and #f(b)# have opposite signs), then there must exist a point #c \in [a,b]# such that #f(c)=0#. Moreover, if we split #[a,b]# in #[a,m] \cup [m,b]#, with #m# being the midpoint of #[a,b]#, then #c# can only lay in one of the two sub-interval. We repeat the test until we have a small enough interval.

In your case, we can observe that #f(-1.5) \approx 1.9 >0#, while #f(-0.5) \approx -0.19 <0#.

So, the solution must be inside #[-1.5, -0.5]#, and this is our first approximation. Of couse we can do better: we can split #[-1.5, -0.5]# into #[-1.5, -1] \cup [-1, -0.5]# and repeat our test. We have #f(-1) = 1>0#. Recapping:

  • #f(-1.5) > 0#
  • #f(-1) > 0#
  • #f(-0.5) < 0#

This means that the function crosses the #x# axis (and thus equals zero) somewhere in #[-1,-0.5]#. This is our second approximation. You can see that this is better than the previous one, since the length of the interval is halved.

We can proceed further and study #f(-0.75) \approx 0.46 > 0# to deduce that the solution is in #[-0.75,-0.5]#.

Usually, when given such an exercise, you are given the starting interval #[a,b]# and are said after how many steps how have to stop. In this example, I solved a hypothetical exercise in which you were given the starting interval #[-1.5,-0.5]# and asked to proceed for three steps.