How to solve a system of logarithmic equations? I need to create a function with the following properties:
$$f(1)=1$$
$$f(65)=75$$
$$f(100)=100$$
Additionally, the function needs to grow logarithmically. So that gives three equations:
$$A \cdot \ln(B \cdot 1 + C) = 1$$
$$A \cdot \ln(B \cdot 65 + C) = 75$$
$$A \cdot \ln(B \cdot 100 + C) = 100$$
I am having trouble with using substitution to solve this. Barring there is no analytic way to solve this system, how would I use a numerical approximation for $A, B$ and $C$?
 A: This is not an answer, just what I tried using Count Iblis's hint. 
Let $$f(x) = A\ln(x+B)+ C$$
You can always write the function with the coefficient of $x$ being $1$, so we assume that it is. 
\begin{align*}
f(1)&=A\ln(1+B)+ C=1\\
f(65)&=A\ln(65+B)+ C=75\\
f(100)&=A\ln(100+B)+ C=100
\end{align*}
We use Count Iblis's strategy (dividing the second and third equations by the first) to get these equations:
\begin{align*}
\frac{75-C}{1-C}&=\frac{\ln(65+B)}{\ln(1+B)}\\
\frac{100-C}{1-C}&=\frac{\ln(100+B)}{\ln(1+B)}
\end{align*}
We can rewrite these as
\begin{align*}
(1+B)^{\frac{75-C}{1-C}} &=65+B\\
(1+B)^{\frac{100-C}{1-C}} &=100 + B
\end{align*}
If we let $m = 1+B$ and $n=1-C$ we get
\begin{align*}
m^{1 + (74/n)} &= m + 64\\
m^{1 + (99/n)} &= m + 99
\end{align*}
Not sure what to do from here ...
A: Hint: Try taking the exp of both sides of each equation.  For instance taking exp of both sides of $A \cdot \ln(B \cdot 1 + C) = 1$ gives $e^{A \cdot \ln(B+ C)}=e.$ That means $e^{\ln(B + C)^A}=e.$
So, $$(B + C)^A=e.$$
