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Math Team Test Question

Date: 30 Jan 1995 23:21:22 -0500
From: Anonymous
Subject: Algebra II

We have been working on this for a week!  A^b - B^a = 10^x.   Given:
A = 1000; B = 1001; with answer X = 3002.99.  Would Dr. Math know how
to factor, manipulate, or tackle this?

Anna-Jonesboro Community High School Math Team.
Anna, Il, 62906.

Any help is VERY much appreciated!
Ruth Roy

Date: 31 Jan 1995 09:28:28 -0500
From: Dr. Ken
Subject: Re: Algebra II


Here's my first question, and I think I know the answer already, but I
always want to make sure.  In your equation, are the two b's the same, and
the two a's the same?  Or do we only know the upper-case A and B?

-Ken "Dr." Math

Date: 31 Jan 1995 20:01:02 -0500
From: Anonymous
Subject: Algebra II help

Thanks for the quick reply... Yes, A = a and B = b... we were trying 
to show the exponent in lower case, as the caret did not look like a caret
We would not know the answer in test situation. We are going over old 
tests to prepare for spring competitions... hope you can give us a new 
way to look at this rascal!

Date: 1 Feb 1995 20:35:19 GMT
From: Dr. Math
Subject: Re: Algebra II

All right.

This is a tricky (and kind of messy) problem. 1000^1001 - 1001^1000 = 10^x, 
or log (base 10) [1000^1001 - 1001^1000] = x.  Basically, this
problem is asking you how many digits are in 1000^1001 - 1001^1000.  

By the way, when I use logs in this message, I'll mean log (base ten)

To see how this works, look at the following example: 893,475,938. It's
a 9-digit number, and it lies between 100,000,000 and 1,000,000,000. The
logs of these two numbers are, respectively, 8 and 9, so the log of
893,475,938 is going to be somewhere between 8 and 9 (it's actually
8.95108).  So the log of any 9-digit number is going to be between 8 and
9, and the log of any n-digit number is going to be between n-1 and n.

Now, I don't really know how I would go about finding an exact value for
x, but I have a pretty good way to approximate it.  First, think about
how many digits are in 1000^1001. It's a 1 with 3003 zeros after it, so
it's a 3004 digit number. It's also _just barely_ a 3004 digit number,
meaning that if you subtract any positive number, you'll get a 3003 or
a number with fewer digits.

Now look at the size of 1001^1000. To do that, we take the log of
1001^1000, and we simplify this to 1000*log[1001].  Now we estimate
log[1001]: it's _just_barely_ over 3, since log[1000] = 3. Using a
calculator (which I hope you can do in your contests), we see that it's
about 3.00043. So our estimate for log [1001^1000] is 1000*3.00043 =
3000.43. So it's a 3001 digit number, and as far as 3001-digit numbers
go, it's about middle-sized.

So what have we done?  We found out that the first number is a 3004-digit
number, and that the second is a 3001-digit number. So what happens when
you subtract them? You're going to get a 3003-digit number, because you
would have had to subtract a 3003-digit number from 1000^1001 to get
anything with fewer digits than 3003.

To see why, try subtracting some numbers from 1000. How big does what you
subtract have to be in order to get anything other than a 3-digit number?

So you know that what you get for x has to be between 3002 and 3003, and
for that matter, that it's a lot closer to 3003 (because what you
subtracted wasn't all that big). That's why the answer was really,
really close to 3003.

But as I was writing this message, a more slick and accurate way occurred
to me.  Recall that the limit as x-> Infinity of [(n+1)/n]^n is e. So the
digits of 1001^1000 are going to be pretty close to the digits of e, at
least in the first couple of places.

Since we know that e is about 2.718, we know that 1001^1000 is about
2.718 x 10^3000. So if we subtract this from 10^3003, we get 
10^3003 - 2.718 x 10^3000 = 10^3000 x (1000 - 2.718)
                          = 10^3000 x (997.282)

And you can take the Log of this number fairly simply with a calculator:
Log(10^3000 x 997.282) = 3000 + Log(997.282)
                       = 3002.9988

That's the deal. If you're like me, you probably looked for a while at
the problem and hoped that it would factor, but no such luck.  It's an
estimation problem.

-Ken "Dr." Math

Date: 06/02/99 at 03:21:04
From: Calla
Subject: Algebra

Dr. Math,

I saw a solution you gave to the question attached. I would like to know 
if my solution is not a little bit more elegant?

Given a^b - b^a =3D 10^x where a = 3D1000, b = 3D1001, prove that
x = 3D3002.998

Now b/a = 3D 1001/1000 = 3D 1.001 so that b = 3D1.001a

a^b - b^a = 3D a^1.001a - (1.001a)^a = 3D 10^x

Divide by a^a   a^0.001a - 1.001^a = 3D (10^x)(a^-a)

1000 - 1.001^1000 = 3D (10^x)(a^-a)

997.2830761 = 3D (10^x)(a^-a)

take logs on both sides    
2.998818449 = 3D x - 1000log(1000) = 3D x - = 3000

x = 3D 3000 + 2.998818449 = 3D 3002.998818

Kind regards

CH vd Westhuizen

Date: 06/02/99 at 17:08:36
From: Doctor Peterson
Subject: Re: Algebra

Hi, Calla.

Yes, the answer you saw is a little more roundabout than it needs to be. 
I'm not sure I'd call your way more elegant; Dr. Ken's solution at the 
end using e is elegant, since it replaces several steps of tedious 
calculation with a surprising insight. But your way is certainly clearer.

With that in mind, I've worked your solution around a little to make it 
as much as possible a straightforward exercise in logs, and here's what 
I have:

    x = log(1000^1001 - 1001^1000)
      = log(1000^1000 * 1000 - (1000*1.001)^1000)
      = log(1000^1000 * (1000 - 1.001^1000))
      = 1000 log(1000) + log(1000 - 10^(1000 log(1.001)))
      = 3000 + log(1000 - 10^0.434)
      = 3000 + log(997.3)
      = 3000 + 2.999 = 3002.999

For comparison, here's a slight reworking of Dr. Ken's final solution:

    e = lim [(n+1)/n]^n

so for large n,

    [(n+1)/n]^n ~ e

Setting n = 1000, this gives approximately

         (1001/1000)^1000 = e

                1001^1000 = e * 1000^1000

    1000^1001 - 1001^1000 = 1000^1001 - e*1000^1000
                          = 1000^1000 * (1000 - e)

    log(1000^1001 - 1001^1000) = log(1000^1000) + log(1000 - e)
                               = 1000*log(1000) + log(997.282)
                               = 3000 + 2.999
                               = 3002.999

Comparing these two solutions, the last part is the same, and the trick 
is to see that

    1.001^1000 = (1 + 1/1000)^1000 ~ e

Since you didn't say how you got 1.001^1000, maybe that's actually what 
you had in mind, rather than what I did explicitly using logs above.

Putting it all together, here's the straightforward AND elegant solution:

    x = log(1000^1001 - 1001^1000)
      = log(1000^1000 * 1000 - (1000*1.001)^1000)
      = log(1000^1000 * (1000 - 1.001^1000))
      = 1000 log(1000) + log(1000 - e)
      = 3000 + log(1000 - 2.718)
      = 3000 + log(997.3)
      = 3000 + 2.999 = 3002.999

Thanks for your suggestion. 

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Basic Algebra

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