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*ln(B*1+C)=1 A*ln(B*65+C)=75 A*ln(B*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?

mriteyl

mriteyl

Answered question

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 ln ( B 1 + C ) = 1
A ln ( B 65 + C ) = 75
A ln ( B 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?

Answer & Explanation

ralharn

ralharn

Beginner2022-09-02Added 15 answers

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.
f ( 1 ) = A ln ( 1 + B ) + C = 1 f ( 65 ) = A ln ( 65 + B ) + C = 75 f ( 100 ) = A ln ( 100 + B ) + C = 100
We use Count Iblis's strategy (dividing the second and third equations by the first) to get these equations:
75 C 1 C = ln ( 65 + B ) ln ( 1 + B ) 100 C 1 C = ln ( 100 + B ) ln ( 1 + B )
We can rewrite these as
( 1 + B ) 75 C 1 C = 65 + B ( 1 + B ) 100 C 1 C = 100 + B
If we let m = 1 + B and n = 1 C we get
m 1 + ( 74 / n ) = m + 64 m 1 + ( 99 / n ) = m + 99
Not sure what to do from here ...

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