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Logarithms Explained and Applied

Date: 10/01/2004 at 11:11:12
From: Tony
Subject: Compute the sum of digits 

What is the number digits contained in the sum of

  2^2001 + 5^2001?

I cannot determine a satisfactory mathematical explanation or 
expression to represent the answer because the number is too large for 
my calculator.

I tried breaking the number into smaller parts and multiplying them, 
i.e., 

  (5^100) * (5^100) * ... [20 times + 1] ... 

but the number simply becomes too large.

Thanks,
Tony


Date: 10/01/2004 at 13:43:30
From: Doctor Ian
Subject: Re: Compute the sum of digits 

Hi Tony,

Looking at first few powers of 2 and 5, 

  2,  4,   8,  16,   32,    64, ...

  5, 25, 125, 625, 3125, 15625, ...

it quickly becomes clear that the 2^2001 term isn't going to matter at
all, as far as the number of digits is concerned.  

So what can we do with 5^2001?  We can use logarithms.  Recall that

   2^3 = 8     <-->    log_2(8) = 3

are just two ways of saying the same thing.  Similarly, 

   10^x = 5    <-->    log_10(5) = x

are two ways of saying the same thing.  That is,

   10^(log_10(5)) = 5

So 

   5^2001 = (10^(log_10(5)))^(2001)

          = 10^(log_10(5)*2001)

Can you take it from here? 

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


Date: 10/01/2004 at 15:13:13
From: Tony
Subject: Compute the sum of digits 

I don't understand what you mean when you say that

  10^x = 5    <-->    log_10(5) = x 

are two ways of saying the same thing.  I'm not familiar with 
logarithms.  Can you explain them, please? 

Thanks,
Tony


Date: 10/01/2004 at 16:12:06
From: Doctor Ian
Subject: Re: Compute the sum of digits 

Hi Tony,

Logarithms are the inverse of exponents, in sort of the same way that
division is the inverse of multiplication.  That is, we can multiply
two numbers to get a third, 

  3 * 48 = 144

or we can start with the product and one of the other factors, and use
division to find the missing factor:

  144 / 3 = 48             144 / 48 = 3

Similarly, we can evaluate an exponent, 

   3
  2  = 2^3 = 2 * 2 * 2 = 8

or we can start with the result and the thing being raised, and use
logarithms to find the missing exponent:

  log (8) = log_2(8) = 3
     2

So 

  2^3 = 8       and       log_2(8) = 3

are just two ways of saying the same thing, in the same way that 

  3 * 4 = 12    and       12 / 3 = 4      and     12 / 4 = 3

are just three ways of saying the same thing.  When we ask

  log_b(a) = x ?          "The logarithm, base b, of a, is what?"

we're asking

  To what power, x, would I have to raise b, to get a?

Here are some examples:

  log_3(81) = 4

  log_10(1,000,000) = 6

  log_5(125) = 3

Let me know if this makes sense to you so far. 

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


Date: 10/05/2004 at 18:11:51
From: Tony
Subject: Compute the sum of digits 

Yes. I understand completely--so far.

Tony


Date: 10/05/2004 at 18:43:42
From: Doctor Ian
Subject: Re: Compute the sum of digits 

Hi Tony,

Excellent.  Next, let's look at a couple of logarithms:

  10^1 =  10         <-->         log_10(10)  = 1

  10^2 = 100         <-->         log_10(100) = 2

Here, I'm using 

    this  <-->  that

to mean "this and that have the same meaning".

Now let's look at one in between:

  10^1 =  10         <-->         log_10(10)  = 1

  10^x =  85         <-->         log_10(85)  = x

  10^2 = 100         <-->         log_10(100) = 2

What can we say about x?  There are lots of things we can say about
it, actually; but for now, the important thing is that we can say that
it's between 1 and 2.  Does that make sense?  

(If x were smaller than one, 10^x would be less than 10.  If it were 
larger than 2, 10^x would be greater than 100.)

In fact, this gives us a way to know how many digits are in a number,
if we know the logarithm (base 10) of the number.  For example, if we
know that 

  log_10(some big number) = 28.5

we know that it's between 10^28 and 10^29, which means it must have 30
digits.  (You can try this on your calculator, if you have a "log"
button.  Type in a number with N digits, and hit "log".  You should
see a number between N-1 and N.  

Anyway, to tie this back to your original problem, if we can make
5^2001 look like 10^(something), then that will tell us how many 
digits it has, without our having to actually figure out what those
digits are. 

Are you still with me?  

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


Date: 10/06/2004 at 08:01:14
From: Tony
Subject: Compute the sum of digits 

I'm still with you...

Tony


Date: 10/06/2004 at 09:05:27
From: Doctor Ian
Subject: Re: Compute the sum of digits 

Hi Tony,

Excellent.  There's just one piece missing from the puzzle, which is
the rule of exponents that 

   (a^b)^c = a^(bc)

For example, 

   (2^3)^4 = (2*2*2)^4

           = (2*2*2)(2*2*2)(2*2*2)(2*2*2)

           = 2^12

           = 2^(3*4)

Does that make sense?  If so, note that there is some x such that

  10^x = 5

This is the same as saying 

  log_10(5) = x

So given

    5^2001

we can rewrite it as

    (10^x)^2001

  = 10^(x*2001)

But what is x?  It's just log_10(5), so we have

  = 10^(log_10(5)*2001)

And what is log_10(5)?  It's about equal to 0.69897.  So we have

  = 10^(0.69897*2001)

  = 10^1398.6

So now we know how many digits are in that number, even if we don't 
know what they are.  

Does it make sense now?

If you _do_ want to know what the digits are, here's the value for
5^2001 returned by Mathematica:

  4354904908 1086083377 8809774738 9436147929 5518713526 9430832174
  6614749144 4267031337 0689236937 5399893940 5327820435 8872756031
  8102151185 9941681625 4139541226 1518430550 7553211551 4865922385
  4561694954 7119240292 8187098617 3595315518 5485855117 0669033412
  0707350364 1181463867 5741973682 8045574038 8682954219 1463592907
  9059744555 6730862845 7835208272 5382878484 8934414823 3312419657
  0054633787 8343282041 4932303510 2391048682 9194537268 3583605554
  2393496585 1333355373 0853318093 7973567503 1959692583 6047603333
  4393391821 5252845658 1377788641 2721596101 1723641499 3116267356
  1491299813 4363082313 4710557529 4992080347 9390045446 0289674834
  6183684569 1367372414 9198586447 9016388335 5778987674 7026356057
  5437108148 9857798248 8140052451 9876725232 8389284772 6713412717
  0926478227 1421214673 1056026697 5687300142 6924577324 8133513449
  4432171071 7824086272 5379213022 7205541778 9537908263 1881353818
  5488987468 0457623242 8674616089 2737356344 5429779608 2701710348
  3581102720 4854352126 7152587391 9071359975 5922270580 1059500996
  6338494815 7918301760 9672949034 6274067433 3179698573 2676361802
  6109468018 1158117248 1511613399 9445116665 8776715097 2907506931
  5638648723 8500289914 0906496083 3766332205 8619772290 4676721858
  2849546541 4874712150 3584526062 0260257995 5322428300 0302014630
  7439405502 3180541147 0822986067 2592763223 5794534189 4753287930
  5088491927 1601460501 9001397161 2697710103 7601110693 3320690236
  2883772710 7695989104 1637968324 2622746115 7464126983 9958113152
  5337696075 439453125

As you can see, there are 1399 digits.  :^D

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


Date: 10/06/2004 at 10:57:50
From: Tony
Subject: Thank you (Compute the sum of digits)

Dr. Ian, your explanation to my problem was thoroughly complete and 
easily understood.

Thank you!
Tony
Associated Topics:
High School Logs

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