Math Team Test QuestionDate: 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 Hello! 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 http://mathforum.org/dr.math/ |
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