Results 1 to 11 of 11

Math Help - logarithmic equation from points

  1. #1
    Newbie
    Joined
    May 2006
    Posts
    12

    logarithmic equation from points

    i need to derive a logarithmic equation in the form y=aloge(x+b)+c
    given that it passes through the points (0,0) (12,20) (30,30)
    it is a positive log with a horizontal asymptote x<0
    can someone please help me work it out cos it is difficult!!!!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by asian5
    i need to derive a logarithmic equation in the form y=aloge(x+b)+c
    given that it passes through the points (0,0) (12,20) (30,30)
    it is a positive log with a horizontal asymptote x<0
    can someone please help me work it out cos it is difficult!!!!!!!
    I presume you need to curve fit (0,0),(12,20),(30,30) such as,
    y=a\ln (x+b)+c
    Subsituting this into this equation yields,
    0=a\ln (0+b)+c
    20=a\ln (12+b)+c
    30=a\ln (30+b)+c
    ---------
    From equation 1 we have,
    c=-a\ln b
    Substitute that into equation 2 and 3 thus,
    20=a\ln (12+b)-a\ln b
    30=a\ln (30+b)-a\ln b
    Thus,
    20=a(\ln (12+b)-\ln b)
    30=a(\ln (30+b)-\ln b)
    Divide these equation thus,
    \frac{2}{3}=\frac{\ln (12+b)-\ln b}{\ln (30+b)-\ln b}
    Cross multiply,
    2\ln (30+b)-2\ln b=3\ln (12+b)-3\ln b
    Now raise e to that power,
    e^{2\ln (30+b)-2\ln b}=e^{3\ln (12+b)-3\ln b}
    Thus,
    \frac{(30+b)^2}{b^2}=\frac{(12+b)^3}{b^3}
    Thus,
    b(30+b)^2=(12+b)^3
    Open sesame,
    30^2b+2\cdot 30b^2+b^3=12^3+3\cdot 12^2b+3\cdot 12 b^2+b^3
    Thus,
    24b^2-468b-1728=0
    Thus,
    b^2-19.5b-72=0
    <to be countinued>
    this gives you b.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    In the previous post I gave you the analytic method, I do not know if you care for that but simply for a solution.
    If you want that then,
    \left{ \begin{array}{c}a\approx 12.8\\ b\approx 3.173\\ c\approx -14.78
    And it gives you this curve below with the points bolded.
    Attached Thumbnails Attached Thumbnails logarithmic equation from points-picture1.gif  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2006
    Posts
    12

    Exclamation finish please

    thanks so much,
    but would u b able to finish the working from the quadratic for "b" u came up with, and sub that value back into the approapriate equation to find the other values. correct to three decimal places
    thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Thus,
    24b^2-468b-1728=0
    Thus,
    b^2-19.5b-72=0
    <to be countinued>
    this gives you b.
    Should be:

    Thus,
    24b^2+468b-1728=0
    Thus,
    b^2+19.5b-72=0

    which has roots b=-22.675, b=3.175, and if we want
    a and c to be real this rules out the first of these
    roots and so b=3.175, which is reassuringly in agreement
    with the numerical result.

    RonL
    Last edited by CaptainBlack; May 28th 2006 at 02:06 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2006
    Posts
    12

    finish

    did you use the quadratic formula?
    what equation do i sub this 'b' value into to give 'a' and 'c'? correct to 3 decimal places?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by asian5
    did you use the quadratic formula?
    what equation do i sub this 'b' value into to give 'a' and 'c'? correct to 3 decimal places?
    Yes I used the quadratic formula.

    Substitute the value of b into any two of the three equations
    given at the top of The PerfectHackers first post, say:

    <br />
0=a\ln (0+b)+c<br />
    <br />
20=a\ln (12+b)+c<br />

    this will leave you with a pair of linear equations in a and b.

    <br />
0=a\ln (3.175)+c<br />
    <br />
20=a\ln (15.175)+c<br />

    or:

    <br />
0=1.1553 a+c<br />
    <br />
20=2.720 a+c<br />
.

    RonL
    Last edited by CaptainBlack; May 28th 2006 at 04:37 AM. Reason: arithmetic error
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2006
    Posts
    12
    now do u solve similtaneously?
    can u take me through it please
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by asian5
    now do u solve similtaneously?
    can u take me through it please
    If you subtarct the first equation from the second:

    <br />
20-0=2.720 a+c-(1.1553 a+c)=1.5644 a<br />

    so:

    <br />
a=20/1.5644 \approx 12.785<br />

    Substituting this back into the first of the equations gives:

    <br />
0=1.1553 a+c=1.1553\times 12.785+c<br />

    so:

    <br />
c=-14.771<br />

    RonL
    Last edited by CaptainBlack; May 28th 2006 at 04:40 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Thank you CaptainBlack for finishing my post, saved me a lot of time.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Thank you CaptainBlack for finishing my post, saved me a lot of time.
    That's OK

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 9th 2011, 05:50 PM
  2. logarithmic equation
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 7th 2009, 09:24 PM
  3. Logarithmic equation
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: September 11th 2009, 01:40 PM
  4. Logarithmic equation
    Posted in the Algebra Forum
    Replies: 4
    Last Post: July 12th 2009, 03:28 PM
  5. Logarithmic equation
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 29th 2008, 07:32 PM

Search Tags


/mathhelpforum @mathhelpforum