Explicit Inverses

Date: 04/21/99 at 02:00:55
From: Sidney Yin
Subject: Solving a variable to a power of itself

Is there a way of solving a question like this:

x^x = 5 

without graphing or guessing?

Thanks.

Date: 04/21/99 at 09:57:51
From: Doctor Mitteldorf
Subject: Re: Solving a variable to a power of itself

Dear Sidney,

In school, you learn to expect that every function has an explicit 
inverse. But this is not usually true. Many more functions don't have 
explicit inverses than do. Write down something as simple as 
x+sin(x) = y and you can't solve for x. Fifth degree polynomials and 
higher don't have explicit inverses.  a^x + b^x = y doesn't have an 
explicit inverse.

The fact that the inverse isn't explicit doesn't really mean very 
much. It just means that the function isn't common enough to have its 
own name. The procedure for solving it isn't necessarily any more 
difficult than the procedure for calculating named functions; for 
example do you have any idea what your calculator has to go through to 
calculate ArcTan(x)to 10 digits?

Here's one way to solve x^x = 5: search for a procedure that you can 
repeat over and over that will get you closer to the answer each time.  
It gives you confidence to start with a procedure that you know will 
stay on the answer once you've found it. So start with the equation 
you have and see if you can work with it:

       x^x = 5        take log of both sides
  x log(x) = log(5)

    x = log(5)/log(x)

Now start with a guess, say 2.1 since we know that 2^2 = 4. The next 
guess is

   x =  log(5)/log(2.1) = 2.169

   Do it again:

   x =  log(5)/log(2.169) = 2.078

Repeating the process gives 2.199, 2.041, 2.255 ...

I was hoping that this procedure would home in on the answer, taking 
us closer and closer each time. But it doesn't - it gets a little 
farther away each time. But notice that each time it jumps across the 
solution: the answers alternate between being too big and being too 
small. This suggests an idea: average them in pairs. In other words, 
if the first guess is 2.1, the second guess won't be log (5)/log(2.1) 
but the average of this number with 2.1. This gives

first guess = 2.1
second guess = 1/2 (log(5)/log(2.1) + 2.1) = 2.1346
third guess = 1/2 (log(5)/log(2.1346) + 2.1346) = 2.1285
fourth guess = 1/2 (log(5)/log(2.1285) + 2.1285) = 2.1295

This is already very close to the solution, 2.12937248276...

You can write a computer program to repeat this process in just a few 
minutes. Even with a programmable calculator, the program you have to 
write is quick and easy - running it 5 or 6 times ("iterations") 
brings you as close as you'll ever need to be to the solution.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/