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Solve It Numerically


Date: 05/21/97 at 11:20:55
From: Nick Riviera
Subject: Exponents and logs

How do you solve the equation x^x = 100?  It doesn't have to be 100, I 
just think that's a nice number. It can be any constant, just as long 
as there isn't a simple integer solution.

I know that the answer is about 3.597285. I have tried everything, 
including taking logs. 

Thanks for your help.


Date: 05/27/97 at 15:29:47
From: Doctor Rob
Subject: Re: Exponents and logs

This equation seldom has an integer solution, never a rational one 
unless it is an integer. Thus, if you can't find an integer solution 
by inspection, you must solve it numerically. There are several ways 
to proceed, of which I will outline two.

Rewrite the equation as x^x - 100 = f(x) = 0. Newton's method (yes, 
Sir Isaac Newton!) says to start with a guess at the answer, say 
x(0) = 4. Then from each guess, you compute a better one using the 
formula:

x(n+1) = x(n) - f(x(n))/f'(x(n)), for every n >= 0

where f'(x) is the derivative of f(x) with respect to x. In your case, 
f'(x) = x^x*(1 + ln(x)), so the equation becomes:

x(n+1) = x(n) - (x^x - 100)/[x^x*(1 + ln(x))]

x(1) = 4 - (4^4 - 100)/[4^4*(1 + ln(4))]
     = 4 - 156/[256*2.38629]
     = 3.74464

This is a better approximation, so compute:

x(2) = 3.620734188
x(3) = 3.597934158
x(4) = 3.597285529
x(5) = 3.597285024
x(6) = 3.597285024

The correct answer is approximately this last value (to 10 significant
figures).

If you don't understand derivatives (a topic in calculus), here is an
alternate approach.

Another iterative method is based on taking logarithms of both sides
of the original equation. Then x*log(x) = log(100). Dividing both 
sides of the equation by log(x), you get:

x = log(100)/log(x).

Now guess x(0) = 4, and use the iteration formula:

x(n+1) = log(100)/log[x(n)]

x(1) = 3.321928095
x(2) = 3.835898515
x(3) = 3.425437005
x(4) = 3.740303368

As you can see, these values are alternately larger and smaller than
the correct one, converging to it but more slowly than before.

x(26) = 3.597892291
x(27) = 3.596810765
...
x(50) = 3.597286641
x(51) = 3.597283760
...
x(75) = 3.597285020
x(76) = 3.597285026
etc.

Even better would be to use the average of x(n) and x(n+1) from the 
latter method:

x(n+1) = (x(n) + log(100)/log[x(n)])/2

Here, with x(0) = 4, we get the answer in 10 steps:

x(10) = 3.597285024

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Calculus

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