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How to Solve Equations with No Analytic Solution Method

Date: 10/26/2005 at 12:02:40
From: Joanne
Subject: x(e^x) -> Simple, short, but how to solve?


One of my friends gave me this question several days ago:
  If x(e^x) = 3, find the value of x.

I realized that even if I changed the base to "ln" form, it would not 
help as the two x's are in different levels:

  x(e^x)= 3
  ln x + x = ln 3
  x = ln 3 - ln x
  x = ln (3/x)

Using "ln", "e", or any sort of combination seems fruitless for this
case.  Please help.

Date: 10/28/2005 at 16:57:52
From: Doctor Fenton
Subject: Re: x(e^x) -> Simple, short, but how to solve?

Hi Joanne,

Thanks for writing to Dr. Math.  Simple-looking problems can be 
difficult to solve.  This equation can't be solved analytically, that 
is, by finding a formula and substituting 3 into it.

Instead, you have to use an iterative method, which makes an initial
estimate ("guess"), and then finds improved estimates by some 

There are many such algorithms, some better (faster) than others.  The 
simplest, which always works when the problem has a solution, but 
which is usually the slowest, is bisection.  You first write the 
problem in the form f(x) = 0, so that you are looking for a root of 
the function f.  Then you need to find two numbers a and b such that 
f(a) and f(b) have different signs.  (I am assuming that f is a 
continuous function, so that it must cross the axis between two values 
where it has different signs.)  Next, you evaluate f at the midpoint 
m=(a+b)/2 of the interval.  If this should happen to be 0, you are 
done.  Otherwise, f(m) will have the same sign as one of the old 
endpoints.  Replace that old endpoint with m, and you have a new 
interval on which f has different signs at the endpoints, and the new
interval is half as large as the original interval.

Then you repeat the above process with the new interval.  After 10 
steps, the interval will be 1/1024 as large as the original interval;
after 20, less than 1/1000000 of the original length.

"False Position" is a method which takes the straight line through
the two points (a,f(a)) and (b,f(b)) and finds where it crosses the
axis, say, at x = c.  Evaluate f(c).  Again, unless f(c) = 0 (possible
but extremely unlikely), you get a new, smaller interval on which
which f has opposite signs at the endpoints.  It generally finds an
approximate root faster than bisection.

The best method is usually Newton's method (although it sometimes 
doesn't work), and it works extremely well here.  It takes calculus to 
derive the formulas, but for this problem, you make an initial 
estimate x_0.  Given an estimate x_n, you make an improved estimate
x_(n+1) by using

                   xe^x - 3
   x_(n+1) = x_n - --------  .

The function f(x) = xe^x - 3, and f(1) < 0, while f(2) > 0 .  Using a 
spreadsheet, with x_0 = 1, I got 10 decimal place accuracy in 5 steps.  
With x_0 = 2, I got 8 decimal place accuracy in 5 steps.

Newton's method generally doubles the number of correct decimals in
each step, once you are "close" to the root.

If you have any questions or need more help, please write back and
show me what you have been able to do, and I will try to offer further

- Doctor Fenton, The Math Forum 

Date: 10/30/2005 at 03:36:39
From: Joanne
Subject: Thank you (x(e^x) -> Simple, short, but how to solve?)

Dear Dr. Fenton,

I would like to express appreciation for the trouble you took to 
explain the methods of solving this question to me.  I have not 
learned such methods yet, but I'm in the process of understanding 
them.  I'll write again should there be anything that I fail to 
understand.  Thank you again.
Associated Topics:
College Algorithms
High School Functions

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