Finding a function that goes through given points

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The discussion revolves around finding a decaying logarithmic function that passes through three specified points while ensuring their sum equals a positive value S. The initial proposed equation is y = -kln(x+1) + S, but there is confusion regarding the choice of constants and the relationship to S. Participants suggest simplifying the equation by keeping certain variables while eliminating others to better fit the logarithmic decay model. After some trial and error, one participant successfully solves the equations by treating A as a variable, leading to a solution. The conversation highlights the complexities of defining the function while adhering to the conditions set by the problem.
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Homework Statement


Find a function of a decaying logarithmic trend that passes through a set of 3 points (at x=1,2,3) whose sum is S, where S>0.

Homework Equations

The Attempt at a Solution


starting point: i assumed the equation had the form: $$y= -kln(x+1)+S$$ with the condition that must satisfy:

$$S=Σ(y(x))$$ from 1→3

but I don't know what to do next
 
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In your proposed generic form, I see no reason to choose x+1 rather than x+an unknown constant. And I don't understand the final S; did you mean this to be the same as the given S or something different?
If we make it x+c as well as a final +d (say) then we have three unknowns (c, d, k) but effectively only one equation. So presumably the question expects a simpler generic form, but it is not clear what. If I had to guess, I would ditch d first.

Edit:After a little background reading, seems like you should keep d and ditch c. Logarithmic decay appears to be defined as the functional inverse of exponential decay, i.e. x=Ae-ky.
 
I see no reason to choose x+1 rather than x+an unknown constant.
Oh, no actually not necessary, I was just trying to find a convenient way to set up the graph on my calculator; my proposed form would have essentially looked like the same graph as that of x=Ae^(-ky)

I don't understand the final S; did you mean this to be the same as the given S or something different?
Same S.I tried your suggested formula. Since the A was left as a free variable to be determined, i had two equations two unknowns and was able to solve. much thanks!
 
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