0.000001 to the Ninety-ninth Power Times Itself

Date: 05/19/99 at 13:46:29
From: Peter and Eben
Subject: 0.000001 to the ninety ninth power times itself

Dear Dr. Math,

We have a question for you that only says error on the calculator. We 
want to know what 0.000001 to the ninety ninth power times 0.000001 to 
the ninety ninth power is.

Thanks, 
Peter and Eben (and Mrs. Bailes)

Date: 05/19/99 at 16:42:09
From: Doctor Peterson
Subject: Re: 0.000001 to the ninety ninth power times itself

Hi, Peter and Eben.

There are some things that are particularly easy for us, that a 
calculator can't handle, because it has a limited number of digits. 
This is one of those cases.

What you want is

            99           99
    0.000001   x 0.000001

which we can write as

    0.000001^99 * 0.000001^99

to make it easier to type.

We can rewrite 0.000001 as 10^-6, using the rule for negative 
exponents. This means that instead of multiplying 1 by 10 six times, 
we have divided 1 by 10 six times. Then the problem becomes

    (10^-6)^99 * (10^-6)^99

We can go one step further: since we are multiplying something by 
itself, we can write that as a square:

    ((10^-6)^99)^2

This means we are raising 10 first to the -6th power, then to the 
99th power, then to the 2nd power.

Now one of the rules of exponents says that

    (a^b)^c = a^(b*c)

This means that if we multiply "b" a's together, and then multiply "c" 
of those together, it's as if we had multiplied "b times c" a's 
together all at once. If you're not familiar with this, try it out on 
a small case like (2^3)^4 and see how it works.

If we apply that rule to our problem, we get

    ((10^-6)^99)^2 = (10^(-6*99))^2 = 10^(-6*99*2) = 10^-1188

That's our answer:

      -1188      1
    10      = ------
                1188
              10

You can write it out as "0." followed by 1187 zeroes and a one. I 
won't bother to write it out; with numbers as large or small as this, 
we usually leave them in this "scientific notation" that lets us see 
how big it is without counting zeroes.

You might be interested in some similar, but harder, problems some 
others have asked about:

 http://mathforum.org/dr.math/problems/pearce12.11.97.html   
 http://mathforum.org/dr.math/problems/sean9.10.97.html   
 http://mathforum.org/dr.math/problems/clifton6.11.98.html   

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