mriteyl

2022-09-01

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:

$$A\cdot \mathrm{ln}(B\cdot 1+C)=1$$

$$A\cdot \mathrm{ln}(B\cdot 65+C)=75$$

$$A\cdot \mathrm{ln}(B\cdot 100+C)=100$$

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 $C$?

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:

$$A\cdot \mathrm{ln}(B\cdot 1+C)=1$$

$$A\cdot \mathrm{ln}(B\cdot 65+C)=75$$

$$A\cdot \mathrm{ln}(B\cdot 100+C)=100$$

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 $C$?

ralharn

Beginner2022-09-02Added 15 answers

This is not an answer, just what I tried using Count Iblis's hint.

Let

$$f(x)=A\mathrm{ln}(x+B)+C$$

You can always write the function with the coefficient of $x$ being $1$, so we assume that it is.

$$\begin{array}{rl}f(1)& =A\mathrm{ln}(1+B)+C=1\\ f(65)& =A\mathrm{ln}(65+B)+C=75\\ f(100)& =A\mathrm{ln}(100+B)+C=100\end{array}$$

We use Count Iblis's strategy (dividing the second and third equations by the first) to get these equations:

$$\begin{array}{rl}\frac{75-C}{1-C}& =\frac{\mathrm{ln}(65+B)}{\mathrm{ln}(1+B)}\\ \frac{100-C}{1-C}& =\frac{\mathrm{ln}(100+B)}{\mathrm{ln}(1+B)}\end{array}$$

We can rewrite these as

$$\begin{array}{rl}(1+B{)}^{\frac{75-C}{1-C}}& =65+B\\ (1+B{)}^{\frac{100-C}{1-C}}& =100+B\end{array}$$

If we let $m=1+B$ and $n=1-C$ we get

$$\begin{array}{rl}{m}^{1+(74/n)}& =m+64\\ {m}^{1+(99/n)}& =m+99\end{array}$$

Not sure what to do from here ...

Let

$$f(x)=A\mathrm{ln}(x+B)+C$$

You can always write the function with the coefficient of $x$ being $1$, so we assume that it is.

$$\begin{array}{rl}f(1)& =A\mathrm{ln}(1+B)+C=1\\ f(65)& =A\mathrm{ln}(65+B)+C=75\\ f(100)& =A\mathrm{ln}(100+B)+C=100\end{array}$$

We use Count Iblis's strategy (dividing the second and third equations by the first) to get these equations:

$$\begin{array}{rl}\frac{75-C}{1-C}& =\frac{\mathrm{ln}(65+B)}{\mathrm{ln}(1+B)}\\ \frac{100-C}{1-C}& =\frac{\mathrm{ln}(100+B)}{\mathrm{ln}(1+B)}\end{array}$$

We can rewrite these as

$$\begin{array}{rl}(1+B{)}^{\frac{75-C}{1-C}}& =65+B\\ (1+B{)}^{\frac{100-C}{1-C}}& =100+B\end{array}$$

If we let $m=1+B$ and $n=1-C$ we get

$$\begin{array}{rl}{m}^{1+(74/n)}& =m+64\\ {m}^{1+(99/n)}& =m+99\end{array}$$

Not sure what to do from here ...

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