2^4 = 16 AND 4^2 = 16

Date: 10/29/2001 at 12:24:38
From: Yvonne
Subject: Exponents

2^4 = 16 AND 4^2 = 16.

Can you think of any other pair of unequal numbers that share the same 
relation as 2 and 4 in the above example? What was your strategy?

Date: 10/29/2001 at 12:31:14
From: Doctor Ian
Subject: Re: Exponents

Hi Yvonne,

My strategy is to think about symmetry, and the properties of even 
exponents. 

Note that 

                1
  (-2)^(-4) = -----
               2^4

What do you suppose (-4)^(-2) is equal to? 

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

Date: 10/29/2001 at 12:51:00
From: Yvonne Ortega
Subject: Re: Exponents

Thank you very much.  I put that down as a possibility, but I thought
perhaps there was another pair of numbers I was not thinking of. Are 
there any other positive numbers you can think of? I sure can't.
Thanks again.

Date: 10/30/2001 at 11:43:39
From: Doctor Ian
Subject: Re: Exponents

Hi Yvonne,

Let's try another strategy.  

You can use the University of Vienna's function plotter at 

  http://www.univie.ac.at/future.media/moe/fplotter/fplotter.html   

to plot the function 

  (2^x) - (x^2)

Now, if you do this, any value of x where the graph crosses the x-axis 
must be a solution to 

  2^x = x^2

and thus the kind of number we're looking for. If you go ahead and 
plot this function, you'll see that the graph crosses the x-axis at 
three points: x = 2 (duh), x = 4 (which we already knew), and at 
approximately -0.7666.  

Let's check that:

  -0.7666^2 = 0.5877

  2^-0.7666 = 0.5878

Which is good enough for government work. Now, I know that you were 
looking for positive numbers... but you were also looking for a 
strategy, and this looks like a reasonable one to explore. 

What happens if we try

  (3^x) - (x^3)

We get a graph that crosses the x-axis at two points: x = 3 (duh), and 
approximately x = 2.478.  Let's check:

  2.478^3 = 15.216
  
  3^2.478 = 15.216

Now, 2.478 is only an approximation, but you can approximate it as 
closely as you like by using something like Newton's Method. And you 
can use the method to find as many pairs of numbers as you like.

Does this help?
  
- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/