logarithmic equation from points

**asian5** · May 27th 2006, 05:49 PM

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!!!!!!!

**ThePerfectHacker** · May 27th 2006, 06:18 PM

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.

**ThePerfectHacker** · May 27th 2006, 06:48 PM

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.

**asian5** · May 28th 2006, 12:27 AM

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

**CaptainBlack** · May 28th 2006, 01:01 AM

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

**asian5** · May 28th 2006, 01:16 AM

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?

**CaptainBlack** · May 28th 2006, 01:33 AM

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:

$ 0=a\ln (0+b)+c $
$ 20=a\ln (12+b)+c $

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

$ 0=a\ln (3.175)+c $
$ 20=a\ln (15.175)+c $

or:

$ 0=1.1553 a+c $
$ 20=2.720 a+c $ .

RonL

**asian5** · May 28th 2006, 01:47 AM

now do u solve similtaneously?
can u take me through it please

**CaptainBlack** · May 28th 2006, 03:35 AM

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:

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

so:

$ a=20/1.5644 \approx 12.785 $

Substituting this back into the first of the equations gives:

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

so:

$ c=-14.771 $

RonL

**ThePerfectHacker** · May 28th 2006, 09:46 AM

Thank you CaptainBlack for finishing my post, saved me a lot of time.

**CaptainBlack** · May 28th 2006, 09:50 AM

Originally Posted by ThePerfectHacker

Thank you CaptainBlack for finishing my post, saved me a lot of time.

That's OK

RonL

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