Curve Fitting

Discussion in 'Other Pre-University Math' started by nycmathguy, Oct 20, 2021.

  1. nycmathguy

    nycmathguy

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    Section 3.3
    Question 83

    Note: I am going to use Example 7 as a guide for me to do 83.

    20211020_031051.jpg

    Question 83

    20211019_035112.jpg

    I have to use the given information in the box above to find at least one point in the form
    (ln x, ln y).

    I will use x = 2, y = 1.189

    So, (ln 1, ln 0.5) becomes (0.693, 0.173).

    I now must find the slope m using (0.693, 0.173) and the origin.

    m = (0.173 - 0)/(0.693 - 0)

    m = 0.173/0.693

    m = 0.2496392496

    Rounding to two decimal places, I get m = 0.25.

    I know use the point-slope formula.

    y - y_1 = m(x - x_1)

    I will use the point (0, 0).

    y - 0 = 0.25(x - 0)

    y = 0.25

    I say the logarithmic equation is y = 0.25 ln x.

    You say?
     
    nycmathguy, Oct 20, 2021
    #1
  2. nycmathguy

    MathLover1

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    why did you use point (0,0)? I do not see figure 3.14.
     
    Last edited: Oct 20, 2021
    MathLover1, Oct 20, 2021
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  3. nycmathguy

    nycmathguy

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    I selected the point (9, 0) following Example 7. As a matter of fact, after checking the answer in the of the textbook, I got it right.

    You say?

    Which two points should I have chosen?
     
    nycmathguy, Oct 20, 2021
    #3
  4. nycmathguy

    MathLover1

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    where is the point (9, 0) in Example 7?(I believe it was typo)

    and you used point (0,0) to get m = (0.173 - 0)/(0.693 - 0)=0.25 which is correct

    but, I do not see fig. 3.14 (in example 7) which they referring to while choosing two points to calculate the slope
     
    Last edited: Oct 20, 2021
    MathLover1, Oct 20, 2021
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  5. nycmathguy

    nycmathguy

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    Here is figure 3.13 and 3.14:

    20211020_221613.jpg
     
    nycmathguy, Oct 21, 2021
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  6. nycmathguy

    MathLover1

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    there you go, now I see
     
    MathLover1, Oct 21, 2021
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  7. nycmathguy

    nycmathguy

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    Cool.
     
    nycmathguy, Oct 21, 2021
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