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Math Help - Homework Logarithm Problem...

  1. #1
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    Homework Logarithm Problem...

    I've gotten into logarithms in school, and it seems interesting to me. I enjoy working on it.

    However, the curriculum I use (homeschool curriculum) isn't clear on a couple points, which I would like to ask about now.

    First: My curriculum gave me a list of common logarithms and antilogarithms. The logarithms on the table range from 1.0 to 9.9 in 0.1 increments, and the corresponding antilogarithms range from .0000 to .9996.

    When the antilogarithm required for an equation is not on the list, I was instructed to round to the closest one (for example, if I'm looking for antilog .4777, log 3.00 is .4771 and log 3.01 is .4786--so I would round to 3.00).

    Well, when I saw the grade on my homework was an F, I couldn't believe it. I thought I did everything correctly. But when I looked at what the correct answers were, I realized what had happened.

    Here's what one of the problems were, and the steps I took to get to the answer. At the end I put their answer to show that it is similar, but not exactly the same (i.e. they rounded their antilog differently than they told me to, I think).

    PROBLEM

    At his son's birth, a man invested $2,000 in savings at 6% for his son's college education. In 19 years, how much, approximately, will be available?

    The formula is A = P(1 + i)^n
    A = accumulated money
    P = principal
    i = interest
    n = number of years

    Substituting the values:

    A = 2000(1.06)^19

    Converting to logarithm (log10 is assumed):

    log A = log 2000 + 19 log 1.06

    Finding the values on the provided table and substituting them in:

    log A = 3.3010 + .4807
    log A = 3.7817

    Convert to antilog to isolate the variable and take out the 3:

    A = (antilog .7817)(10^3)

    According to their table, log 6.05 is .7818 and log 6.04 is .7810. 6.05 is closer, so I used that value and multiplied by 10^3 to find A:

    A = 6,050

    The accumulated amount of money over 19 years will be $6,050

    My answer was given an F. I looked at their answer: $6,049. Obviously they had used a different value for antilog .7817.

    I figured out that in order to get $6,049 as the answer, antilog .7817 would have to be 6.049. But how would one figure out to use this logarithm?

    Second question, which is really just the first one taken a bit further...

    PROBLEM

    (Using the value of A from the previous problem)

    Because of inflation, money is depreciating in real value. The formula for the real value P of A dollars n years from now at the rate of i per year is P = A(1 + i )^-n. If the rate of inflation is 8% a year, what is the current real value of the A = 2,000(1.06)19 dollars from the preceding problem above? After 19 years, did the father and son have more or less money, as measured by current real value, than they did at the son's birth?

    Using that formula, I can break it down after a few steps to this:

    log P = 3.1472
    P = (antilog .1472)(10^3)

    There is no exact match in the table. 1.40 is .1461 and 1.41 is .1492. 1.40 is closer, so using that, I get the answer:

    P = 1400
    $1,400 is the real value of the saved money

    Their answer is $1,403. In order to get to this answer, antilog .1472 would have to be 1.403...how could I possibly guess that?

    Thanks for taking the time to read this huge problem, I hope I made it clear what the problem is and I hope you can help me! Please and thank you!
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  2. #2
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    "...I was instructed to round to the closest one ."

    You followed that but they gave you F because they did not follow that.
    That is your problem, or, that is the problem.

    Forgetting that one above...them not following rules..., their answers are following the correct procedure.
    They are interpolating, or using proportion, to get to the correct answer. They are not rounding off.

    -------
    A = (antilog .7817)(10^3)

    According to their table, log 6.05 is .7818 and log 6.04 is .7810. 6.05 is closer, so I used that value and multiplied by 10^3 to find A:

    A = 6,050

    The accumulated amount of money over 19 years will be $6,050

    My answer was given an F. I looked at their answer: $6,049. Obviously they had used a different value for antilog .7817.

    I figured out that in order to get $6,049 as the answer, antilog .7817 would have to be 6.049. But how would one figure out to use this logarithm?


    Your 0.7817 falls between 0.7810 and 0.7818. So your antilog should fall between 6.04 and 6.05
    Difference between 0.7810 and 0.7818 is 0.0008
    Diffference between 0.7810 and your 0.7817 is 0.0007
    Difference between 6.04 and 6.05 is 0.01
    So, you interpolate or solve by proportion. Let c = correction,
    c/0.0007 = 0.01/0.0008
    c = (0.01)*(0.0007/0.0008) = 0.009
    That then should be added to the 6.04.
    Hence, your 0.7817 should have an antilog of 6.04 +0.009 = 6.049

    -----------------------------------------
    "log P = 3.1472
    P = (antilog .1472)(10^3)

    There is no exact match in the table. 1.40 is .1461 and 1.41 is .1492. 1.40 is closer, so using that, I get the answer:

    P = 1400
    $1,400 is the real value of the saved money

    Their answer is $1,403. In order to get to this answer, antilog .1472 would have to be 1.403...how could I possibly guess that?"


    Same thing. Your 0.1472 falls between 0.1461 and 0.1492. So the antilog of your 0.1472 should fall between 1.40 and 1.41

    Difference of 0.1461 and 0.1492 is 0.0031
    Difference of 0.1461 and 0.1472 is 0.0011
    Difference of 1.40 and 1.41 is 0.01
    So,
    c/0.0011 = 0.01/0.0031
    c = (0.01)(0.0011/0.0031) = 0.00355
    Therefore, antilog of 0.1472 = 1.40 +0.0036 = 1.4036
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  3. #3
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    THANKS

    Thank you so much. I assumed they weren't rounding off, but I didn't know the formula to find the correction.

    I will put this to good use right away!

    Thanks again, you rock.
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  4. #4
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    Another problem :(

    Hey, thanks for helping me with the logarithms. Can I ask one more question?

    I'm trying to use your correction formula. Here's how I have it set up:

    PROBLEM: find antilog .6660
    log 4.63 = .6656
    log 4.64 = .6665

    difference between .6656 and .6660 = .0004
    difference between .6656 and .6665 = .0009
    difference between 4.63 and 4.64 = .01

    c/.0004 = .01/.0009
    c = .0004 * .01 / .0009
    c = .0044...45

    add it to 4.63 = 4.6344...45

    so the answer should be 4.6345, right? But they wanted the answer to be 4.639. Why, do you think?
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  5. #5
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    Quote Originally Posted by shirkdeio View Post
    Hey, thanks for helping me with the logarithms. Can I ask one more question?

    I'm trying to use your correction formula. Here's how I have it set up:

    PROBLEM: find antilog .6660
    log 4.63 = .6656
    log 4.64 = .6665

    difference between .6656 and .6660 = .0004
    difference between .6656 and .6665 = .0009
    difference between 4.63 and 4.64 = .01

    c/.0004 = .01/.0009
    c = .0004 * .01 / .0009
    c = .0044...45

    add it to 4.63 = 4.6344...45

    so the answer should be 4.6345, right? But they wanted the answer to be 4.639. Why, do you think?
    Your interpolation is correct. Antilog(.6660)=4.6345
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  6. #6
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    Thanks again

    Okay, just making sure I did it right. Thanks.
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