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!!!!!!!

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- May 27th 2006, 05:49 PMasian5logarithmic 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!!!!!!! - May 27th 2006, 06:18 PMThePerfectHackerQuote:

Originally Posted by**asian5**

$\displaystyle y=a\ln (x+b)+c$

Subsituting this into this equation yields,

$\displaystyle 0=a\ln (0+b)+c$

$\displaystyle 20=a\ln (12+b)+c$

$\displaystyle 30=a\ln (30+b)+c$

---------

From equation 1 we have,

$\displaystyle c=-a\ln b$

Substitute that into equation 2 and 3 thus,

$\displaystyle 20=a\ln (12+b)-a\ln b$

$\displaystyle 30=a\ln (30+b)-a\ln b$

Thus,

$\displaystyle 20=a(\ln (12+b)-\ln b)$

$\displaystyle 30=a(\ln (30+b)-\ln b)$

Divide these equation thus,

$\displaystyle \frac{2}{3}=\frac{\ln (12+b)-\ln b}{\ln (30+b)-\ln b}$

Cross multiply,

$\displaystyle 2\ln (30+b)-2\ln b=3\ln (12+b)-3\ln b$

Now raise $\displaystyle e$ to that power,

$\displaystyle e^{2\ln (30+b)-2\ln b}=e^{3\ln (12+b)-3\ln b}$

Thus,

$\displaystyle \frac{(30+b)^2}{b^2}=\frac{(12+b)^3}{b^3}$

Thus,

$\displaystyle b(30+b)^2=(12+b)^3$

Open sesame,

$\displaystyle 30^2b+2\cdot 30b^2+b^3=12^3+3\cdot 12^2b+3\cdot 12 b^2+b^3$

Thus,

$\displaystyle 24b^2-468b-1728=0$

Thus,

$\displaystyle b^2-19.5b-72=0$

<to be countinued>

this gives you b. - May 27th 2006, 06:48 PMThePerfectHacker
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,

$\displaystyle \left{ \begin{array}{c}a\approx 12.8\\ b\approx 3.173\\ c\approx -14.78$

And it gives you this curve below with the points bolded. - May 28th 2006, 12:27 AMasian5finish 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 - May 28th 2006, 01:01 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

Thus,

$\displaystyle 24b^2+468b-1728=0$

Thus,

$\displaystyle b^2+19.5b-72=0$

which has roots $\displaystyle b=-22.675, b=3.175$, and if we want

$\displaystyle a$ and $\displaystyle c$ to be real this rules out the first of these

roots and so $\displaystyle b=3.175$, which is reassuringly in agreement

with the numerical result.

RonL - May 28th 2006, 01:16 AMasian5finish
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? - May 28th 2006, 01:33 AMCaptainBlackQuote:

Originally Posted by**asian5**

Substitute the value of $\displaystyle b$ into any two of the three equations

given at the top of The PerfectHackers first post, say:

$\displaystyle

0=a\ln (0+b)+c

$

$\displaystyle

20=a\ln (12+b)+c

$

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

$\displaystyle

0=a\ln (3.175)+c

$

$\displaystyle

20=a\ln (15.175)+c

$

or:

$\displaystyle

0=1.1553 a+c

$

$\displaystyle

20=2.720 a+c

$.

RonL - May 28th 2006, 01:47 AMasian5
now do u solve similtaneously?

can u take me through it please - May 28th 2006, 03:35 AMCaptainBlackQuote:

Originally Posted by**asian5**

$\displaystyle

20-0=2.720 a+c-(1.1553 a+c)=1.5644 a

$

so:

$\displaystyle

a=20/1.5644 \approx 12.785

$

Substituting this back into the first of the equations gives:

$\displaystyle

0=1.1553 a+c=1.1553\times 12.785+c

$

so:

$\displaystyle

c=-14.771

$

RonL - May 28th 2006, 09:46 AMThePerfectHacker
Thank you CaptainBlack for finishing my post, saved me a lot of time.

- May 28th 2006, 09:50 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

RonL