decimal numbers as power of 10

Discussion in 'Math' started by davidray, Dec 17, 2009.

  1. davidray

    Thread Starter New Member

    Dec 17, 2009
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    How do you express the decimal numbers like 10,000 & 1,000,000 as a power of 10 ???
     
  2. eng1ne

    Member

    Dec 4, 2009
    97
    3
    10 000 = 10^4

    1 000 000 = 10^6

    Count the zeroes, look at the power; notice the pattern?
     
  3. Papabravo

    Expert

    Feb 24, 2006
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    You can express any number at all as a power of 10 using logarithms. In the case of powers of ten, the logarithm happens to be an integer. It is also the case that exponetiation and taking the logarithm are inverse functions of each other.
     
  4. zgozvrm

    Member

    Oct 24, 2009
    115
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    To find what power of 10 equals a specific number X (any real number), you have 10^Y = X. Take the base 10 log of both sides of the equation: \log{10^Y} = \log{X}. Then using \log{(10^A)} = A, we have: Y = \log{X}.

    So, for X= 5.296, for instance, we have Y = \log{(5.296)} = 0.72394797643163

    Then, you'll find that 10^{0.72394797643163} = 5.296
     
  5. eng1ne

    Member

    Dec 4, 2009
    97
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    Apologies for my simple reply! I misinterpreted the question.
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    I don't think so -- your reply was just fine for the OPs question. We were just trying to point out the much richer structure of logarithms and exponents as an additional FYI for the curious among us.

    There is also the possible ulterior motive of demonstrating our extended body of knowledge and experience. Do 'ya think?
     
  7. Ratch

    New Member

    Mar 20, 2007
    1,068
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    Papabravo,

    But not negative numbers, unless you want to tangle with complex numbers.

    zgozvrm,

    You can be more general than that. Suppose I want the binary logarithm of 1024. Do ln(1024)/ln(2) = 6.9314718056/0.69314718056 = 10 . Or use a different log base for calculation; log(1024)/log(2) = 3.0102995664/0.301029995664 = 10 . But don't try to find the logarithm of a negative number unless you can accept complex numbers.

    Ratch
     
  8. zgozvrm

    Member

    Oct 24, 2009
    115
    2

    Of course, you can use any logarithm (natural, base 10, base 7, etc.), but since we're already trying to solve for 10^x, it just makes sense to use the base 10 logarithm; it simplifies the calculation slightly.
     
  9. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    For negative numbers you just put a minus sign in front of both sides as in
    Code ( (Unknown Language)):
    1.  
    2.  y =  10^x
    3. -y = -10^x
    4.  
    Where is the need for compex numbers?

    They do come about if you want to say that there is a Laurent series that conveges to ln(z), that is analytic in some neighborhood, except for a removeable singularity at the origin. There is also the matter of restricting arg(z) so tha the function does not become multivalued.
     
  10. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Papabravo,

    What is log10(-3)? How are you going to put a minus sign in front of that?

    Answer is the complex number (0.47712125472,1.36437635384). In other words, 10^(0.47712125472,1.36437635384) = -3 .

    Ratch
     
  11. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    We are talking about two different problems. The original problem was to represent a real number as a power of 10.

    The second problem is to take the logarithm of a negative number. It is a related problem and your exposition is correct. I have no argument with anything that you've said. In the case of trying to take the inverse of the equation -y = -10^x, order of operations is crucial, and the operations are NOT commutative. This state of affairs happens all the time.
     
  12. Ratch

    New Member

    Mar 20, 2007
    1,068
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    Papabrave,

    Yes, but negative numbers are real numbers, too. I was just being pedantic in pointing out that logs of negative numbers are not real like positive numbers are.

    Ratch
     
  13. Harrington

    New Member

    Dec 19, 2009
    86
    3
    I read your articles here with interest and so I quickly set about writing you a very tiny peace of software to make life easier using the Java specification

    If you can run java then here is a small program attached

    Have fun with this for those of you who would like to see how we do this here I've also attached the code

    I'm using a single frame launched from main.java

    A couple of textfields which pass a double value to the variables then the Math class to get my answers when buttons receive events I've also included the JOption class so that in the event of an error been thrown the textfields are reset to "" The process can be repeated any number of times

    To run the application simply open a dos box after downloading the attachments to a folder of your choice CD or Change Directory to that folder using the dos commands and then type

    java -jar Logarithms.jar Up should come a window and away you go

    Enjoy
     
    Last edited: Mar 3, 2010
  14. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
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    Thanks Harrington. That makes life a little 'quicker'.
     
  15. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
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    Before calculators, there was a standard reference book - 'Logarithmic and Other Tables' - that's one way of finding the log of any number?

    (Or, even less accurately, a Slide Rule) :rolleyes:

    Still got both...
     
  16. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    For folks who went to college before calculators were invented, most of you will no doubt remember what a pain it was to have to do e.g. calculations to 5 or more significant figures using log tables -- or, even worse, by long-hand techniques.

    For a humbling experience, set yourself up a reasonably complicated problem including multiplications, divisions, roots/powers, and some trig functions. Then figure out the answer using e.g. an old CRC math tables book. Oh, and time yourself too.

    No fair using a calculator or computer, either. You have to get the answer with those old methods and be confident of the result.

    Every decade or so I'll try such a task and marvel at how the old skills have (nearly) completely atrophied. I suppose we could relearn the task if necessary, but it's horrible drudgery that no one would want to go back to.

    I wonder if anyone younger than their 50's learned such things -- did they continue to teach such skills after calculators came out?
     
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