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Log Components and Mantissas

  1. Dec 13, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    Show how 10^(-1.7) = alog(0.3) * 10^(-2)

    2. Relevant equations

    Component: non-fraction part of a logarithm.
    Mantissa: fraction part of a logarithm.

    3. The attempt at a solution

    I'm "reverse engineering" a prof's class through watching his lectures, so I don't have the benefits of having his self-published book sitting in front of me, office hours, or TAs to ask for assistance.

    I am completely lost by how he achieves this conversion. Any hints on how to understand this stuff about logs and mantissas, etc?
     
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  3. Dec 13, 2013 #2

    Dick

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    The whole discussion of mantissas etc. dates back to the days when you did this stuff by looking values up in printed tables, but ok. 10^(-1.7)=10^(0.3)*10^(-2). 10^(0.3) is what you call alog(0.3). To do this in the proper spirit of the exercise I think you ought to find a printed table of logarithms and look up 0.3 as a value of the base 10 logarithm and conclude that it's the log of approximately 2. But maybe that's a little too antiquarian. I'm surprised they are still teaching this in the calculator age.
     
    Last edited: Dec 13, 2013
  4. Dec 13, 2013 #3

    Qube

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    Well, he was a TA at my school in 1967, earned his doctorate in 1977, and has been teaching for the last 40 years, so I'd imagine some things are a bit engrained. Also to note, this isn't a mathematics course; this is a chemistry course. This happens to be an application of mathematics in chem. He violently opposes any use of calculators when one can use his or her "noggin." As a matter of fact, he says he'd horsewhip any teacher who allowed students to use calculators and upon the second offense, he'd hang the teacher, so I wouldn't put it across him.

    The above tangent aside:

    I'm not seeing how he transformed 10^(-1.7) to the logarithm, however. He said something about subtracting 1 from -1, and countering that by adding one to -0.7. After that he got -2.3.

    From here he derived the logarithm. I think I see what's going on from the number -2.3 onward:

    0.3: mantissa.
    -2: component.

    log (mantissa) * 10^(component)

    But why does this work (if the above is even correct)? Also what's up with the adding one and subtracting one?

    https://scontent-b-iad.xx.fbcdn.net/hphotos-prn2/v/1475833_10201260848904985_2098432365_n.jpg?oh=e45b11375ea22a41736ede751dd9e7f1&oe=52ADE679
     
    Last edited: Dec 13, 2013
  5. Dec 13, 2013 #4

    Dick

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    It's really unnecessary to add and subtract 1 and I don't see what -2.3 has to do with anything. You want to express -1.7 as the sum of a number between 0 and 1 (that's the mantissa) and an integer (that's the component). -1.7=0.3+(-2). 0.3 is the mantissa and -2 is the component. End of story.
     
  6. Dec 13, 2013 #5

    Qube

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    That's really clear and just the explanation I was searching for! Thanks!

    Oops, log(0.3) * 10^-2 doesn't seem to = 10^-1.7. What's up?
     
  7. Dec 13, 2013 #6

    Dick

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    Did your prof write the image you posted, or did you? It's amazingly awful. For your 'Oops', no, it's alog(0.3)*10^(-2). alog(0.3) is the antilog of 0.3, the number whose log is 0.3. That's 10^(0.3).
     
  8. Dec 13, 2013 #7

    Qube

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    Oh I transcribed what he wrote on the board. I was unaware that alog referred to antilog. That makes more sense!
     
  9. Dec 13, 2013 #8

    Dick

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    He didn't use the phrase "young whippersnappers"? Goes well with "horsewhip".
     
  10. Dec 13, 2013 #9

    Qube

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    Nope, but I assure you, he knows all the chemistry vocabulary there is. As a matter of fact, he knows more about general chemistry than anyone else on this planet.

    http://youtu.be/NOnvrkpnkoI?t=23m33s
     
  11. Dec 13, 2013 #10

    Dick

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    I'm really going to hope you transcribed that a little wrong because it came out a mess. Don't worry about it. Just concentrate on rules of logs and exponents. That's what you need to know once you get out of the course. You'll probably never hear about 'mantissas' ever again.
     
  12. Dec 13, 2013 #11

    Dick

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    Ok, then take the chemistry. Forget mantissas. Chemistry is hard enough.
     
  13. Dec 13, 2013 #12

    Qube

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    I hope so. For now we're expected to know log values off the top of our heads, such as log 2 which equals 0.3010. No log tables provided. He doesn't want us to use a calculator to know what 10^(0.3) equals. This is going to be a long semester, but I chose this prof for a reason (I wanted a challenge, and right now I'm trying to cover as much ground as possible before he comes around and steamrolls me with chemistry).

    Yeah, I admit, I "transcribed" it. Most of that it just additions I made. While we're at this, I feel as if I should do anyone else who comes across this thread a favor. First, it's the characteristic of a log, not component. That was my first mistake in "transcribing" the prof's words. Mantissa is the correct term however.

    Second, here are a few rules I found regarding logs, characteristics, and mantissas. A very clear expansion on your excellent explanation.

    http://books.google.com/books?id=aGyBPu75gi4C&pg=PA18&dq=mantissa+characteristic&hl=en&sa=X&ei=K-6rUqOzHMWpkAetnoDwBQ&ved=0CC8Q6AEwAA#v=onepage&q=mantissa%20characteristic&f=false [Broken]

    Third, alog isn't a variable a multiplied by the log of some argument. It's the antilog. This was my "oops."

    Fourth, if you have a professor this stuck up about logs, then you (reader of this PF thread in the future) should probably also realize there are only as many significant figures in a log as there are in the mantissa (the fraction or decimal part). So the log of say 0.00000928 only has 3 significant figures.

    I think that about covers it for PF patrons, past, present, and future!
     
    Last edited by a moderator: May 6, 2017
  14. Dec 13, 2013 #13

    Dick

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    Good advice. Are you allowed to xerox (or mimeograph) a table of logs or use a slide rule? I remember log(2) is about 0.301, but that's just because I'm pretty old myself and I used to do that. I don't see the point. Being able say log(2) is around 1/3 is one thing. Knowing a more accurate value is an abuse of the "noggin" when you are doing chemistry. You need that part of your mind to memorize a thousand other things.
     
    Last edited by a moderator: May 6, 2017
  15. Dec 15, 2013 #14

    Qube

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    I don't think we're allowed to bring in any extra materials to tests.

    He's correlated America's decline in various measures of math literacy to the rise in calculator usage.

    So, we're not allowed to use calculators; any teacher who allows calculator usage is to be publicly shamed or exterminated; and we're supposed to know the log of 2 (and probably 3, 4, 5, 6, 7, and so on).

    Eccentricities aside, I find this guy an excellent teacher. Sure, he might say something that sounds like Martian to me one day, but after some rumination, it begins to make sense. I sort of appreciate this kind of teaching as opposed to the kind of teaching that puts a plate of facts in front of you for you to memorize without further thought.

    Whether my GPA will like this teacher is another issue, but that's why I'm desperately trying to cram in all the information he has to offer now, before the spring semester starts. If this were a marathon, then I'm trying to get to the 20 mile mark before anyone else even toes the start line.

    Liking a teacher doesn't get you into med school. GPA does. So far I have a 4.0 (thanks to the great people at Physics Forums in no small part!)
     
    Last edited: Dec 15, 2013
  16. Dec 15, 2013 #15
    As an avid runner, I really liked this statement! You would win for sure unless you "ran" slower than a brisk walk (over 20:00 per mile). :biggrin:

    I learned something new from this topic, even if it is *mostly* useless now that calculators dominate.
     
  17. Dec 15, 2013 #16

    Qube

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    I'm an avid runner too! Hence the metaphor :). Great to see that you learned something. Maybe you'll come to my university and put it to use with this prof!
     
  18. Dec 15, 2013 #17

    Dick

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    As far as the integers go, if you know log(2), then it's easy to find log(4)=log(2^2) and log(8)=log(2^3). And if you know log(2) and log(3) then you can also easily find log(6)=log(2*3) and log(9)=log(3^2). You can also deduce log(5)=log(10/2). That just leaves you with log(2), log(3) and log(7) to hard memorize, if you don't mind doing some simple math to find the other ones.
     
    Last edited: Dec 15, 2013
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