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I need to find the logarithmic curve between two points $$A(0,5),\quad B(180,9)$$

We know that the formula for logarithmic function is: $\;f(x) = \log(x)\,\;$so $$ 5 = \log(0),\quad 9 = \log(180)$$

But that's impossible because $\log(0)$ is undefined. What Did I do wrong?

Following the below advice I'm still stuck

$a^5=0-b$ and $a^9=180-b$

then

$$a^9 = 180+a^5 $$ $$a^4 = 180$$ $$a = 3.66$$

Now let's plug a in our original formula

$$3.66^5 = -b$$ $$b = -656.7$$

$$f(x) = \log3.66(x+656.7)$$

I did a little bit of fiddling with a graph and at the end of the day what I was looking for was

$$f(x) = \log1.77(x-5)$$

I would be awesome to understand how to achieve this result without playing randomly with excel.

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  • $\begingroup$ Do you have to fit a logarithm curve? If yes try to find a,b, c so that $f(x)=a.\log(x/c+b)$ and go through A, B $\endgroup$ – Katsu Oct 28 '13 at 12:57
  • $\begingroup$ you mean f(x)=a*log(x/c+b)? $\endgroup$ – Kahel Oct 28 '13 at 13:37
  • $\begingroup$ yes it's what I wrote. '.' is same as 'x' or '*' in my country. $\endgroup$ – Katsu Oct 28 '13 at 14:38
  • $\begingroup$ $\log1.77(x-5)$ is undefined at $x=0$ $\endgroup$ – RE60K Mar 10 '15 at 11:47
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Assume that $f(x)=log(ax+10)+c$

We have $f(0)=5$, so

$f(0)=log(0x+10)+c=5 $

So $c$ is $4$. Now we know that:

$f(x)=log(ax+10)+4$

We have $f(180)=9$, so

$f(180)=log(180a+10)+4=9$

that is,

$log(180a+10)=5$

$a=(10^5-10)/180$

hence $f(x)$ is (you can do the required simplification if you want to...):

$f(x) = log((10^5-10)/180)x+10)+4 $

Needless to say that there may be other functions that could be logarithmic and can pass by A, B.

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Try to use $f(x) = \log{(a(x-b)})$. Then try find the values of a and b.

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  • $\begingroup$ edited but still in the dark :( $\endgroup$ – Kahel Oct 28 '13 at 14:12
  • $\begingroup$ Ok. To continue, you should get the expression a^9 = 180 + a^4. This is a polynomial equation that cannot be reduced algebraically though. $\endgroup$ – aaa Oct 28 '13 at 14:17
  • $\begingroup$ Edit. I think I got it. . . $\endgroup$ – Kahel Oct 28 '13 at 14:53
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Please try

$$ y = A \log B x $$

There are 2 constants and 2 equations.

It is easier to work with

$$ x = a e^ {b x } $$

and then to get back logs.

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