Log passes through two points

Question

LogK (X+A)+B passes through 0,0 and 3,5 (or as close to 0 as log can get). How would one find these values, given the required B changes with K

Answer 1

There seem to be different interpretations of this question.

(1) Assuming the expression is log(K(X+A))+B:

Let y=log(K(X+A))+B, and B=pK where p is a constant.

y=log(K(X+A))+pK.

(0,0): 0=log(KA)+pK;

(3,5): 5=log(3K+KA)+pK.

So, 5=log(3K+KA)-log(KA)=log((3K+KA)/KA)=log(3/A+1).

3/A+1=10⁵ or e⁵ depending on the log base being 10 or e. Assume base e (natural log).

3/A=e⁵-1, A=3/(e⁵-1).

-pK=ln(3K/(e⁵-1)).

If K=1, p=-ln(3/(e⁵-1))=ln((e⁵-1)/3). 

p=B/K=ln((e⁵-1)/3)/K, B=ln((e⁵-1)/3). So we found A and B, but we can't find K, because all we know is that B changes with K so that B/K is a constant.

(2) Assuming the expression is log_K(X+A)+B: let y=log_K(X+A)+B.

(0,0): 0=log_K(A)+B, B=log_K(1/A);

(3,5): 5=log_K(3+A)+B.

5=log_K(3/A+1), K⁵=3/A+1, 3/A=K⁵-1, A=3/(K⁵-1), B=log_K((K⁵-1)/3).