mriteyl

2022-09-01

How to solve a system of logarithmic equations?
I need to create a function with the following properties:
$f\left(1\right)=1$
$f\left(65\right)=75$
$f\left(100\right)=100$
Additionally, the function needs to grow logarithmically. So that gives three equations:
$A\cdot \mathrm{ln}\left(B\cdot 1+C\right)=1$
$A\cdot \mathrm{ln}\left(B\cdot 65+C\right)=75$
$A\cdot \mathrm{ln}\left(B\cdot 100+C\right)=100$
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 $C$?

ralharn

This is not an answer, just what I tried using Count Iblis's hint.
Let
$f\left(x\right)=A\mathrm{ln}\left(x+B\right)+C$
You can always write the function with the coefficient of $x$ being $1$, so we assume that it is.
$\begin{array}{rl}f\left(1\right)& =A\mathrm{ln}\left(1+B\right)+C=1\\ f\left(65\right)& =A\mathrm{ln}\left(65+B\right)+C=75\\ f\left(100\right)& =A\mathrm{ln}\left(100+B\right)+C=100\end{array}$
We use Count Iblis's strategy (dividing the second and third equations by the first) to get these equations:
$\begin{array}{rl}\frac{75-C}{1-C}& =\frac{\mathrm{ln}\left(65+B\right)}{\mathrm{ln}\left(1+B\right)}\\ \frac{100-C}{1-C}& =\frac{\mathrm{ln}\left(100+B\right)}{\mathrm{ln}\left(1+B\right)}\end{array}$
We can rewrite these as
$\begin{array}{rl}\left(1+B{\right)}^{\frac{75-C}{1-C}}& =65+B\\ \left(1+B{\right)}^{\frac{100-C}{1-C}}& =100+B\end{array}$
If we let $m=1+B$ and $n=1-C$ we get
$\begin{array}{rl}{m}^{1+\left(74/n\right)}& =m+64\\ {m}^{1+\left(99/n\right)}& =m+99\end{array}$
Not sure what to do from here ...

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