I need to create a logarithmic equation with 3 points known. I need to create a logarithmic equation that intersects with the points:
$(1,504), (20,803.25), (500,7526.925)$
It should be something like this: $$y=\ln(x)+504$$
There must be some extra variables so that the graph hits the above points but I don't know what they would be or where they would be.
 A: Oops -
my answer was wrong!!!!!
If the equation is
$y = af(x)+b$,
then it can only be fit to
$(x_i, y_i)_{i=1}^3$
when
certain conditions hold.
$y_i = a\ln(x_i)+b$
for $i = 1, 2$.
Subtracting,
$y_2-y_1
=a(f(x_2)-f(x_1))
$
so
$a
=\frac{y_2-y_1}{f(x_2)-f(x_1)}
$.
Applying this to
$i = 1,3$,
we have
$a
=\frac{y_3-y_1}{f(x_3)-f(x_1)}
$.
Therefore,
to have this fit,
it is necessary to have
$\frac{y_2-y_1}{f(x_2)-f(x_1)}
=\frac{y_3-y_1}{f(x_3)-f(x_1)}
$.
Since this does not hold
for the given points
when
$f(x) = \ln(x)$,
no fit of this kind is possible.
Therefore a three-parameter fit
seems to be necessary.
A: This is a partial answer where I explore two possible forms of $f(x)$ :
Assuming $y = f(x) = a\ln(bx)+c$. We define $B := \ln(b)$
$$
\begin{cases}
a\ln(b)+c=504 \\
a\ln(20b) + c = 803.25 \\
a\ln(500b) + c = 7526.925
\end{cases} \implies \begin{cases}
aB+c=504 \\
a\ln(20) + aB+ c = 803.25 \\
a\ln(500) + aB+ c = 7526.925
\end{cases} \implies \begin{cases}
a\ln(20) + aB+ 504-aB= 803.25 \\
a\ln(500) + aB+ 504-aB = 7526.925
\end{cases} \implies \begin{cases}
a\ln(20) = 299.25 \\
a\ln(500) = 7022.925
\end{cases} \implies \begin{cases}
a = 99.8921 \\
a = 1130.55
\end{cases}
$$
So we have an incompatibility here 
Assuming $y = f(x) = a\ln(x) + b$
$$
\begin{cases}
a\ln(1)+b=504 \\
a\ln(20)+b=803.25 \\
a\ln(500)+b=7526.925
\end{cases} \implies b = 504 \implies \begin{cases}
a\ln(20) = 299.25 \\
a\ln(500) = 7022.925
\end{cases}
$$ which we have already seen that it is incompatible.
